Chapter 14: Dual Nature of Radiation and Matter

In 1887, a young German physicist named Heinrich Hertz was busy proving that light is a wave. His apparatus was simple in idea: two metal spheres separated by a small air gap, with a high voltage across them. When a spark jumped the gap, it sent out electromagnetic waves, and a second gap some distance away sparked in response. Hertz was confirming Maxwell’s prediction that electromagnetic waves exist and travel through space — a triumph for the wave picture of light.

While adjusting the set-up, Hertz noticed something he had not gone looking for. When ultraviolet light was allowed to fall on the metal electrodes, the spark jumped across far more easily. Shield the electrodes from that light, and the spark became reluctant again. Light shining on a metal was somehow helping electricity escape from it.

Hertz recorded the observation and moved on. He could not have guessed that this small side-effect would refuse to fit into the very wave theory he had just confirmed. Over the next twenty years it would force physicists to accept an uncomfortable idea: that light travels as a wave, but arrives as a particle — and that matter, in turn, hides a wave inside it.

Diagram

Fig. 14.0 – Hertz’s spark-gap set-up: two metal spheres with a spark across the gap, ultraviolet light falling on the electrodes, and electrons leaving the metal surface.

NoteCuriosity Corner

Q1. Why does a metal give out electrons when light of one colour falls on it, but stays completely silent under another colour — however bright that light is made? Q2. When light strikes a metal, why do electrons come out at once, within a billionth of a second, instead of taking time to soak up enough energy? Q3. What exactly is a “photon”, and how much energy does a single photon of ordinary light carry? Q4. How does a photocell switch street lights on at dusk and off at dawn, all by itself? Q5. If light can behave like a stream of particles, can an electron behave like a wave? Q6. Every moving object is said to have a wavelength. Then why do we never notice the wave nature of a running student or a car on a highway?

By the end of this chapter, you will be able to answer each of these questions yourself.

14.1 Introduction

Before starting this chapter, check whether you can answer the following from your earlier study: 1. What is electromagnetic radiation? 2. What are the characteristics of a wave? 3. What do you mean by frequency and wave number associated with a wave? 4. What are the characteristic properties of particles of matter? 5. How do we define momentum of a particle? 6. What are the different types of energies that a particle of matter can possess?

The Wave Picture of Light

In earlier chapters you have studied several optical phenomena — reflection, refraction, interference, diffraction and polarization. Light is electromagnetic radiation, and almost all of these phenomena have been explained by treating light as a wave.

Two of them are worth pausing on. Interference happens when two light waves overlap and add up or cancel out, and diffraction happens when light bends around an obstacle or spreads after passing through a narrow slit. Only something that travels as a wave can behave this way, so these phenomena are strong evidence for the wave nature of light.

Visible light, however, is only a narrow band of a much larger family. You are already familiar with the wave nature of electromagnetic radiation in other regions of the spectrum as well — X-rays, γ-rays, infrared and ultraviolet radiation, and microwaves. All of these are the same kind of radiation; they differ only in frequency and wavelength.

The different regions of the electromagnetic spectrum are used in daily life exactly because their frequencies differ. Microwaves carry mobile-phone and satellite signals, infrared radiation is used in remote controls and thermal imaging, and X-rays pass through soft tissue to image bone. The physics of the wave is the same in each case; only the frequency changes.

Structurally, electromagnetic radiation consists of mutually perpendicular oscillating electric and magnetic fields, and both these fields are perpendicular to the direction in which the wave and its energy are travelling, as shown in Fig. 14.1a.

Diagram

Fig. 14.1a – An electromagnetic wave showing the oscillating electric field and magnetic field in mutually perpendicular planes, both perpendicular to the direction of propagation.

NoteDefinition

Electromagnetic radiation is radiation made up of mutually perpendicular oscillating electric and magnetic fields, both of which are perpendicular to the direction in which the wave and its energy travel.

Where the Wave Picture Begins to Fail

In Chapter 3, on Kinetic Theory of Gases and Radiation, you came across the spectrum of black body radiation. A black body is an ideal body that absorbs all the radiation falling on it, and when heated it emits radiation whose spread over different wavelengths depends only on its temperature.

This spectrum cannot be explained using the wave nature of radiation. The wave theory predicts a curve of one shape; the measured curve has a quite different shape, as suggested in Fig. 14.1b.

Diagram

Fig. 14.1b – Black body radiation curve: intensity versus wavelength, with the experimentally observed curve and the (failing) prediction of wave theory drawn together.

Such phenomena appear during the interaction of radiation with matter, and they need quantum physics to explain them. That is the common thread running through this entire chapter: light behaves like a wave while it travels, but something else happens when it meets matter.

Q. Does this mean the wave theory of light is wrong?

A. No. It is incomplete, not wrong. Interference, diffraction and polarization are still explained perfectly by the wave theory. It simply cannot handle a second set of observations, and for those we need a new picture.

Planck’s Idea: Energy Comes in Packets

The idea of quantization of energy was first proposed by Planck to explain the black body spectrum. To quantize a quantity means to allow it only in whole-number multiples of a fixed basic amount, instead of in any amount you please.

Planck proposed a model that says (i) energy is emitted in packets, and (ii) at higher frequencies, the energy of a packet is large.

Planck assumed that atoms behave like tiny oscillators that emit electromagnetic radiation only in discrete packets,

\[E = nh\nu\]

where \(\nu\) is the frequency of the oscillator, \(h\) is a constant (its value is given in Section 14.2.4), \(n\) is a whole number, and \(E\) is the energy emitted. In SI units, \(E\) is in joule (J), \(\nu\) is in hertz (Hz, that is \(\mathrm{s^{-1}}\)), and \(h\) is in joule second (J s).

The emissions occur only when the oscillator makes a jump from one quantized level of energy to another of lower energy. The oscillator cannot slide down smoothly; it can only drop from one allowed level to a lower allowed level, releasing the energy difference as a packet of radiation.

NotePrinciple / Law

Planck’s quantization hypothesis: an atomic oscillator of frequency \(\nu\) emits electromagnetic radiation only in discrete packets of energy given by \(E = nh\nu\), and emission takes place only when the oscillator jumps from one quantized energy level to a lower one.

Think of the difference between a ramp and a staircase. On a ramp you can stop at any height you like; on a staircase you can only stand on one step or another. Classical physics expected the energy of an oscillator to behave like a ramp. Planck proposed a staircase — and only the staircase gave the correct black body curve.

Planck introduced this idea around the year 1900, and he introduced it reluctantly. He was trying to fit the measured black body curve, and quantization was the only assumption that made his formula match the data. For several years even Planck treated it as a mathematical device rather than a statement about how nature really works.

Notice what the second part of Planck’s model implies: since a packet at higher frequency carries more energy, radiation of high frequency delivers its energy to matter in large lumps, while low-frequency radiation delivers it in small lumps. This single idea decides almost everything that follows in this chapter.

This model of Planck turned out to be the basis for Einstein’s theory to explain the observations of experiments on the photoelectric effect, which we will study in the following section.

Q. In \(E = nh\nu\), what does the whole number \(n\) physically mean?

A. It tells you how many packets of energy are involved. Since \(n\) can only be \(1, 2, 3, \dots\), the energy given out can be one packet, two packets and so on — but never one and a half packets. The size of each packet is fixed by the frequency.

In examination questions, this section usually appears indirectly: you are expected to know that the energy of a packet is proportional to frequency, so a statement such as “increasing the intensity increases the energy of each packet” is false. Intensity and frequency control two different things, and keeping them apart is the key skill for the whole chapter.

14.2 The Photoelectric Effect

How the Effect Was First Noticed

Heinrich Hertz discovered photoelectric emission in 1887, while he was working on the production of electromagnetic waves by spark discharge. He noticed that when ultraviolet light is incident on a metal electrode, a high voltage spark passes across the electrodes.

What was actually happening was this: electrons were being emitted from the metal surface. These free electrons in the gap made it easier for the spark to pass, which is why the spark improved whenever ultraviolet light fell on the electrode. Hertz saw the effect of the emission before anyone understood the emission itself.

A surface that emits electrons when it is illuminated with appropriate radiation is known as a photosensitive surface.

NoteDefinition

A photosensitive surface is a surface that emits electrons when radiation of a suitable kind falls on it.

What the Effect Is

The phenomenon of emission of electrons from a metal surface, when radiation of appropriate frequency is incident on it, is known as the photoelectric effect. The process is shown in Fig. 14.1, where radiation arriving at the metal surface is followed by electrons leaving that surface.

Diagram

Fig. 14.1 – Process of photoelectric effect: ultraviolet radiation falling on a metal plate and electrons being ejected from its surface.

NoteDefinition

The photoelectric effect is the emission of electrons from a metal surface when radiation of appropriate frequency is incident on it.

Read the definition carefully. It says radiation of appropriate frequency, not radiation of appropriate brightness. This is the first hint that frequency, and not intensity, is the deciding factor — a point that the experiments in the next sections will establish beyond doubt.

Which Metals Respond to Which Radiation

The radiation that works is not the same for every metal. For metals like zinc, cadmium and magnesium, ultraviolet radiation is necessary, while for alkali metals, even visible radiation is sufficient.

Alkali metals are the metals of the first group of the periodic table — sodium, potassium, caesium and their neighbours. They hold their outermost electrons rather loosely, which is why ordinary visible light is enough to knock electrons out of them, whereas zinc or cadmium demands the higher frequency of ultraviolet radiation.

Q. Why should the type of metal decide which radiation is required, if the electrons emitted are the same ordinary electrons in every case?

A. Because the electrons are not equally tightly held. Every metal has its own minimum energy that must be supplied before an electron can escape from its surface. A metal that holds its electrons tightly needs the more energetic ultraviolet radiation, while a metal that holds them loosely can be worked with visible light. This minimum energy is given a name and a symbol in Section 14.2.3.

Electrical energy can be obtained from light (electromagnetic radiation) in two ways: (i) the photo-emissive effect, described above, in which electrons are actually thrown out of a metal surface, and (ii) the photo-voltaic effect, used in a solar cell. In the latter case, an electrical potential difference is generated in a semiconductor using solar energy, and no electron leaves the material.

In examinations, a common one-mark trap is a question asking whether red light can produce photoelectric emission from a given metal. The answer never depends on how bright the red light is; it depends only on whether that metal responds to visible radiation at all.

14.2.1 Experimental Set-up of Photoelectric Effect

The Apparatus

A typical laboratory experimental set-up for the photoelectric effect is shown in Fig. 14.2. It consists of an evacuated glass tube with a quartz window, containing a photosensitive metal plate — the emitter E — and another metal plate — the collector C.

Diagram

Fig. 14.2 – Schematic of experimental set-up for photoelectric effect: evacuated tube with quartz window, emitter E and collector C, variable voltage source with a voltmeter, ammeter in the circuit and light source S.

Two features of the tube deserve attention. The tube is evacuated so that the emitted electrons can travel from E to C without colliding with air molecules and losing energy on the way. The window is made of quartz rather than ordinary glass because ordinary glass absorbs ultraviolet radiation, while quartz allows it through to reach the emitter.

The emitter and collector are connected to a voltage source whose voltage can be changed, and to an ammeter to measure the current in the circuit. A potential difference of \(V\), as measured by the voltmeter, is maintained between the emitter E (the cathode) and the collector C (the anode), normally C being at a positive potential with respect to the emitter.

Controlling the Potential Difference

This potential difference can be varied, and C can even be at a negative potential with respect to E. The circuit is therefore able to do two opposite jobs, simply by reversing the polarity of the source, as indicated in Fig. 14.2a.

Diagram

Fig. 14.2a – The same circuit drawn twice, once with the collector positive (electrons pulled towards C) and once with the collector negative (electrons pushed back towards E).

When the anode potential \(V\) is positive, it accelerates the electrons, and it is therefore called the accelerating potential. When the anode potential \(V\) is negative, it retards the flow of electrons, and it is therefore known as the retarding potential.

NoteDefinition

Accelerating potential is the potential difference applied with the collector positive with respect to the emitter, so that emitted electrons are pulled towards the collector.

NoteDefinition

Retarding potential is the potential difference applied with the collector negative with respect to the emitter, so that emitted electrons are opposed on their way to the collector.

The ability to reverse the polarity is what makes this apparatus so useful. With a positive collector, the experimenter can count how many electrons are being emitted, because eventually every emitted electron is collected. With a negative collector, the experimenter can measure how energetic the electrons are, because only those with enough kinetic energy manage to fight their way across.

Q. Why is the tube evacuated instead of being simply closed with air inside?

A. Electrons leaving the emitter are very light and are easily deflected. In air they would collide with gas molecules, lose energy and scatter in all directions, so the current reaching the collector would depend on the air pressure rather than on the light. In vacuum, the only things acting on an electron are the applied electric field and the energy it started with.

The Light Source

A source S of monochromatic light (light corresponding to only one specific frequency) of sufficiently high frequency, that is short wavelength \(\lambda \le 10^{-7}\) m, is used.

Monochromatic light is essential here. If the beam contained many frequencies together, each frequency would produce electrons of a different energy, and it would be impossible to say which frequency produced which result. By fixing the frequency, the experimenter can change one quantity at a time — intensity, or frequency, or voltage — and watch what happens.

What Happens Inside

Light is made to fall on the surface of the metal plate E, and electrons are ejected from the metal through its surface. These electrons, called photoelectrons, are collected at the collector C.

NoteDefinition

Photoelectrons are ordinary electrons emitted from a metal surface as a result of radiation falling on it; they are given this special name only to indicate that they are emitted due to incident light.

We now know that free electrons are available in a metal plate. They are emitted if sufficient energy is supplied to them to overcome the barrier that keeps them inside the metal.

That word “barrier” is important. A free electron inside a metal can wander from atom to atom quite easily, which is why metals conduct, but it is still trapped inside the metal as a whole. Something must supply it with enough energy to cross the surface and get out. In this experiment, that something is the incident light.

In the late nineteenth century, these facts were not known. Scientists working on the photoelectric effect performed various experiments with set-ups like this one and simply noted down what they saw. Their observations are summarized in the next section, and we will then try to analyze the observations and their explanation.

14.2.2 Observations from Experiments on Photoelectric Effect

What the Experimenters Could Change

Before reading the observations, it helps to be clear about the three things an experimenter can adjust in the set-up of Fig. 14.2, because every observation below is the result of changing one of them and keeping the others fixed.

The first is the intensity of the incident radiation, which means the energy falling on unit area of the emitter in unit time — in everyday language, how bright the light is. The second is the frequency of the incident radiation, which is fixed by the source and decides the “colour” or the region of the spectrum. The third is the potential difference \(V\) applied between emitter and collector, which can be accelerating or retarding.

Q. Is a brighter light the same thing as a higher-frequency light?

A. No, and keeping the two apart is the single most important habit in this chapter. Intensity tells you how much energy arrives per second on unit area. Frequency tells you what kind of radiation it is. A very bright red lamp and a very dim ultraviolet lamp can be arranged to deliver the same intensity, yet they behave completely differently in this experiment.

The quantity measured in every case is the photocurrent \(I\), read on the ammeter. Since each photoelectron carries the same charge, the photocurrent is a direct measure of the number of electrons reaching the collector per second.

Frequency Decides Whether Emission Happens at All

Observation 1. When ultraviolet radiation was incident on the emitter plate, current \(I\) was recorded even if the intensity of radiation was very low. Photocurrent \(I\) was observed only if the frequency of the incident radiation was more than some threshold frequency \(\nu_0\). This \(\nu_0\) was the same for a given metal and was different for different metals used as the emitter. For a given frequency \(\nu\) (\(> \nu_0\)) of the incident radiation, no matter how feeble the light was — meaning however small the intensity of radiation may be — electrons were always emitted.

NoteDefinition

Threshold frequency \(\nu_0\) is the minimum frequency of incident radiation below which no photoelectrons are emitted from a given metal surface, however intense the radiation may be. Its value depends only on the metal used as the emitter.

This is a surprising result, and it is worth stating in the two directions separately. Below \(\nu_0\), no amount of brightness helps: the current stays at zero. Above \(\nu_0\), no amount of dimness hurts: some current always flows.

Observation 2. There was no time lag between the incidence of light and the emission of electrons. The photocurrent started instantaneously — within \(10^{-9}\) s — on shining the radiation, even if the intensity of radiation was low. As soon as the incident radiation was stopped, the flow of current stopped.

Notice that emission begins and ends with the light itself. The metal does not “store up” light energy and release electrons afterwards.

Intensity Decides How Many Electrons Come Out

Observation 3. Keeping the frequency \(\nu\) of the incident radiation and the accelerating potential \(V\) fixed, if the intensity was increased, the photocurrent increased linearly with intensity, as shown in Fig. 14.3.

Diagram

Fig. 14.3 – Photocurrent as a function of incident intensity for fixed incident frequency and accelerating potential: a straight line through the origin.

A straight line through the origin means simple proportionality: double the intensity and you double the photocurrent, that is, double the number of electrons emitted per second.

Observation 4. The photocurrent \(I\) could also be varied by changing the potential of the collector plate. For given incident radiation (intensity and frequency both fixed), \(I\) depended on the accelerating potential \(V\), that is, on the potential difference between the emitter and the collector. Initially the current increased with voltage, but then it remained constant. This constant value was termed the saturation current \(I_0\), as seen in Fig. 14.4. The superscripts 1, 2, 3 of \(I_0\) in the figure refer to different intensities of the incident light.

Diagram

Fig. 14.4 – Photocurrent as a function of accelerating potential for fixed incident frequency and three different incident intensities, showing saturation currents \(I_0^1\), \(I_0^2\), \(I_0^3\) and a common stopping potential \(-V_0\).

NoteDefinition

Saturation current \(I_0\) is the maximum photocurrent obtained for a given incident radiation, reached when a further increase in the accelerating potential produces no further increase in the current.

Observation 5. Keeping the accelerating voltage and the incident frequency fixed, if the intensity of incident radiation was increased, the value of the saturation current also increased proportionately. For example, if the intensity was doubled, the saturation current was also doubled.

In a photoelectric experiment with a fixed frequency of incident light and a fixed accelerating potential, the saturation current is \(12\ \mu\)A when the intensity of the source is \(2 \times 10^{-5}\ \mathrm{W\,m^{-2}}\). What will be the saturation current if the intensity is raised to \(5 \times 10^{-5}\ \mathrm{W\,m^{-2}}\), the frequency remaining unchanged?

What Decides the Energy of the Emitted Electrons

Observation 6. The maximum kinetic energy \(KE_{max}\) of the electrons, and hence their maximum velocity, depended on the potential \(V\) for a given metal used for the emitter plate and for a given frequency of the incident radiation. If the material was changed, or the frequency of the incident radiation was changed, \(KE_{max}\) changed. It did not depend on the intensity of the incident radiation. Thus, even for a very small incident intensity, if the frequency of incident radiation was larger than the threshold frequency \(\nu_0\), the value of \(KE_{max}\) from a given surface was always the same for a given incident frequency.

Here the word maximum matters. Not every emitted electron has the same energy; \(KE_{max}\) refers to the most energetic electron leaving the surface. The reason for this spread is taken up later in the chapter.

Observation 7. If increasingly negative potentials were applied to the collector, the photocurrent decreased, and for some typical value \(-V_0\) the photocurrent became zero. \(V_0\) was termed the cut-off or stopping potential. The fact that some current still flowed before this value was reached indicated that when the potential was retarding, the photoelectrons still had enough energy to overcome the retarding (opposing) electric field and reach the collector. The value of \(V_0\) was the same for any incident intensity as long as the incident frequency was the same, as seen in Fig. 14.4, but was different for different emitter materials.

NoteDefinition

Stopping potential (or cut-off potential) \(V_0\) is the magnitude of the retarding potential at which the photocurrent becomes exactly zero for a given emitter material and a given frequency of incident radiation.

Q. Why is the stopping potential such a useful thing to measure?

A. Because it converts an energy measurement into a voltage measurement. When the current has just fallen to zero, even the fastest electron has been brought to a halt at the collector. The retarding voltage needed to do that therefore tells us how much kinetic energy that fastest electron started with — and a voltmeter is far easier to use than any direct measurement of electron speed.

Changing the Frequency Changes the Stopping Potential

Observation 8. If the frequency of incident radiation was changed keeping the intensity constant, then the saturation current remained the same but the stopping potential \(V_0\) changed. In Fig. 14.5, \(V_0^1\), \(V_0^2\) and \(V_0^3\) are the stopping potentials for incident frequencies \(\nu_1\), \(\nu_2\) and \(\nu_3\) respectively, with \(\nu_3 > \nu_2 > \nu_1\).

Diagram

Fig. 14.5 – Photocurrent as a function of accelerating potential for fixed incident intensity but three different incident frequencies on the same emitter material, showing a common saturation current and three different stopping potentials.

The stopping potential \(V_0\) varied linearly with \(\nu\), as shown in Fig. 14.6. For different metals, the slopes of such straight lines were the same, but the intercepts on the frequency axis and on the stopping potential axis were different.

Diagram

Fig. 14.6 – Stopping potential as a function of frequency of incident radiation for emitters made of different metals: parallel straight lines with different intercepts.

Read Fig. 14.6 carefully, because it carries two separate messages. Parallel lines mean that the rate at which stopping potential grows with frequency is a property of nature, the same for every metal. Different intercepts mean that where each line begins — the threshold frequency — is a property of the particular metal.

Observation 9. The photocurrent, and hence the number of electrons, depended on the intensity but not on the frequency of the incident radiation, as long as the incident frequency was larger than the threshold frequency \(\nu_0\) and the potential of the anode was higher than that of the cathode.

The Pattern Behind the Nine Observations

Taken together, these observations divide neatly into two groups, and this division is what examination questions test again and again.

Intensity controls the number of photoelectrons, and therefore the photocurrent and the saturation current. Frequency controls the energy of the photoelectrons, and therefore the stopping potential, and it also decides whether there is any emission at all.

Not one of these observations mentions any waiting time, and none of them allows brightness to substitute for frequency. As the next section shows, this is precisely where the wave theory of light runs into trouble.

14.2.3 Failure of Wave Theory to Explain the Observations from Experiments on Photoelectric Effect

The Problem of Instantaneous Emission

Most of the observations listed in the previous section could not be explained by the wave theory of electromagnetic radiation. First and foremost was the instantaneous emission of electrons on the incidence of light.

The wave picture would expect the metal surface to absorb the incident energy continuously, the way a beach absorbs the energy of waves arriving on it. All the electrons near the surface would go on absorbing energy little by little. The metal surface would require a reasonable time — of the order of a few minutes to a few hours — to accumulate sufficient energy to knock off electrons.

The same picture also predicts a clear dependence on brightness. The greater the intensity of incident radiation, the more will be the incident energy, and hence the shorter the expected time required to knock off the electrons. For small incident intensity, the energy incident on unit area in unit time will be small, and it will take longer to knock off the electrons.

These arguments were contradictory to the observations. In the actual experiment, emission began within \(10^{-9}\) s, and it began even when the light was extremely feeble.

Let us try to estimate the time that will be required for the photocurrent to start, so that we can see just how badly the wave picture fails. We need to define the term ‘work function’ of a metal for this exercise.

The Work Function of a Metal

We know that metals have free electrons. This fact makes metals good conductors of heat and electricity. These electrons are free to move inside the metal but are otherwise confined inside the metal. They cannot escape from the surface unless sufficient energy is supplied to them, as suggested in Fig. 14.6a.

Diagram

Fig. 14.6a – An electron free to move inside a metal but held back at the surface by an energy barrier, with the minimum energy needed to cross the barrier marked as the work function.

The minimum amount of energy required to be provided to an electron to pull it out of the metal from the surface is called the work function of the metal and is denoted by \(\phi_0\).

NoteDefinition

The work function \(\phi_0\) of a metal is the minimum amount of energy that must be supplied to an electron at the surface of that metal to pull it out of the metal.

Work function depends on the properties of the metal and on the nature of its surface. Values of the work function of metals are generally expressed in a unit of energy called the electron volt (eV).

One electron volt is the energy gained by an electron when it is accelerated through a potential difference of one volt, so \(1\ \mathrm{eV} = 1.6 \times 10^{-19}\) J. This unit is used because the joule is far too large a unit for a single electron; a work function of a few electron volts is a comfortable number to write and to compare across metals.

You have studied the ionization energy of an atom — the energy needed to remove an electron from an isolated atom. What ionization energy is to an atom, the work function is to a solid, which is a large collection of atoms.

Table 14.1 gives typical values of the work function for some common metals.

Table 14.1: Typical values of work function for some common metals.

Metal Work function (in eV)
Potassium 2.3
Sodium 2.4
Calcium 2.9
Zinc 3.6
Silver 4.3
Aluminum 4.3
Tungsten 4.5
Copper 4.7
Nickel 5.0
Gold 5.1

Q. Why do the alkali metals sit at the top of this table with the smallest work functions?

A. Because they hold their outermost electrons most loosely. Potassium needs only 2.3 eV to release an electron from its surface, while gold needs 5.1 eV — more than twice as much. This is exactly why a cell meant to work with ordinary visible light uses an alkali metal for its cathode and never gold or nickel.

Estimating the Waiting Time Predicted by Wave Theory

We can now put numbers to the wave theory’s prediction. The idea is simple: work out how much energy the incident light delivers to a single electron per second, and then find how long that electron must wait before it has collected an amount equal to the work function.

Radiation of intensity \(0.5 \times 10^{-4}\) W/m² falls on the emitter in a photoelectric set-up. The emitter (cathode) is made of potassium and has an area of 5 cm². Assume that only the electrons on the surface are knocked off by the radiation. According to wave theory, how much time will be required before some deflection is noticed in the microammeter connected in the circuit? (Given: the metallic radius of a potassium atom is 230 pm, and the work function of potassium is 2.3 eV.)

Given

Intensity of radiation \(= 0.5 \times 10^{-4}\) W/m²,

Area of cathode \(= 5\ \mathrm{cm^2} = 5 \times 10^{-4}\ \mathrm{m^2}\),

Radius of potassium atom \(= 230\ \mathrm{pm} = 230 \times 10^{-12}\) m,

Work function of potassium \(= 2.3\ \mathrm{eV} = 2.3 \times 1.6 \times 10^{-19}\) J.

The number \(N\) of electrons present on the surface of the cathode can be approximately calculated by assuming that each potassium atom contributes one electron and that the radius of a potassium atom is \(230 \times 10^{-12}\) m. Dividing the area of the cathode by the area covered by one atom gives the number of atoms on the surface, and hence the number of surface electrons.

\[N = \frac{\text{Area of cathode}}{\text{Area covered by one atom}} = \frac{5 \times 10^{-4}}{3.1415 \times 230 \times 10^{-12} \times 230 \times 10^{-12}} = 3008 \times 10^{12}\]

Incident power on the cathode is the intensity multiplied by the area on which the radiation falls:

\[P = 0.5 \times 10^{-4}\ \mathrm{W/m^2} \times 5 \times 10^{-4}\ \mathrm{m^2} = 2.5 \times 10^{-8}\ \mathrm{W}\]

Wave theory assumes that this power, distributed over the whole area of the cathode, is uniformly absorbed by all the electrons. This assumption is the heart of the calculation: every surface electron gets an equal share. Therefore the energy absorbed by each electron in one second is

\[\frac{2.5 \times 10^{-8}\ \mathrm{W}}{3009 \times 10^{12}} \approx 8.311 \times 10^{-24}\ \mathrm{W}\]

Work function of potassium is

\[2.30\ \mathrm{eV} = 2.30 \times 1.6 \times 10^{-19}\ \mathrm{J} = 3.68 \times 10^{-19}\ \mathrm{J}\]

Hence each electron will require a minimum of \(3.68 \times 10^{-19}\) J of energy to be knocked off from the surface of the cathode.

The time required to accumulate this energy will be the energy needed divided by the energy collected per second:

\[t = \frac{3.68 \times 10^{-19}\ \mathrm{J}}{8.31 \times 10^{-24}\ \mathrm{W}} = 4.428 \times 10^{4}\ \mathrm{s}\]

which is about half a day.

Q. What exactly does this half-a-day answer prove?

A. It proves that the wave theory is not slightly off but wrong by an enormous factor. The experiment shows emission starting within \(10^{-9}\) s, while the wave calculation predicts about \(4.4 \times 10^{4}\) s. That is a mismatch of roughly thirteen orders of magnitude, and no small refinement of the wave picture can close such a gap.

Two More Failures

Secondly, since a larger incident intensity implies larger energy, the electrons are expected to be emitted with larger kinetic energy. But the observation showed that the maximum kinetic energy did not depend on the incident intensity and depended instead on the incident frequency. According to wave theory, the frequency of incident radiation has no role at all in determining the kinetic energy of photoelectrons.

Moreover, wave theory expected photoelectrons to be emitted for any frequency, provided the intensity of radiation was large enough. But the observations indicated that for a given metal surface, some characteristic cut-off frequency \(\nu_0\) existed, below which no photoelectrons were emitted however intense the incident radiation was; and photoelectrons were always emitted if the incident frequency \(\nu\) was greater than \(\nu_0\), even if the intensity was low.

Three failures, then, and all of them of the same kind. The wave theory keeps insisting that energy is delivered smoothly and shared out among many electrons, while the experiment keeps insisting that energy is delivered to one electron at a time, in a lump whose size is fixed by the frequency. The next section takes that hint seriously.

14.2.4 Einstein’s Postulate of Quantization of Energy and the Photoelectric Equation

Einstein’s Bold Extension of Planck’s Idea

Planck’s hypothesis of energy quantization, which he had used to explain black body radiation, was extended by Einstein in 1905 to all types of electromagnetic radiation.

The step Einstein took was bigger than it first looks. Planck had quantized only the emission by the atomic oscillators; the radiation itself was still thought of as a spreading wave. Einstein proposed that the radiation stays in lumps while it travels, and delivers itself to matter in lumps as well.

Einstein proposed that under certain conditions, light behaves as if it was a particle, and its energy is released or absorbed in bundles or quanta. He named the quantum of energy of light a photon, with energy \(E = h\nu\), where \(\nu\) is the frequency of light and \(h\) is the constant defined by Planck in his model to explain black body radiation. It is now known as Planck’s constant and has a value \(6.626 \times 10^{-34}\) J s.

\[E = h\nu \qquad \text{--- (14.1)}\]

Here \(E\) is the energy of one photon in joule (J), \(\nu\) is the frequency of the radiation in hertz (Hz), and \(h\) is Planck’s constant in joule second (J s).

NoteDefinition

A photon is a quantum, or single indivisible bundle, of energy of electromagnetic radiation, carrying energy \(E = h\nu\) where \(\nu\) is the frequency of the radiation.

It may be noted that Equation (14.1) is a relation between a particle-like property, the energy \(E\), and a wave-like property, the frequency \(\nu\). Equation (14.1) is known as Einstein’s relation.

Q. Why is it strange to have \(E\) and \(\nu\) in the same equation?

A. Because in classical physics these two words belong to different worlds. Energy of a particle is something you assign to a single, localized object. Frequency is something only a spread-out, repeating wave can have — a single grain of sand has no frequency. Equation (14.1) says that light insists on having both at once, and that is precisely the “dual nature” named in the chapter’s title.

Energy, Frequency and Wavelength

Einstein’s relation (14.1) holds good for the entire electromagnetic spectrum. It says that the energy of electromagnetic radiation is directly proportional to the frequency, and, since \(\nu = c/\lambda\), it is inversely proportional to the wavelength. Here \(c\) is the speed of electromagnetic radiation in vacuum and \(\lambda\) is the wavelength in metre (m).

Hence high frequency radiation means high energy radiation. Alternatively, short wavelength radiation means high energy radiation.

This single sentence explains a great deal that you already know from elsewhere. γ-rays and X-rays, at the short-wavelength end, are the ones that can damage tissue and are handled with shielding; radio waves, at the long-wavelength end, pass through us all day without any effect at all. The difference is not how much total energy arrives, but how big each individual bundle is.

Q3. What exactly is a “photon”, and how much energy does a single photon of ordinary light carry?

A. A photon is one indivisible packet of energy of electromagnetic radiation, and its energy is \(E = h\nu\) — fixed entirely by the frequency of the radiation. For visible light, a single photon carries only a few electron volts of energy, that is, a few times \(10^{-19}\) J. That is an extremely small amount on the human scale, which is why an ordinary lamp sends out enormous numbers of photons every second and we never notice them arriving one by one.

(a) Calculate the energies of photons corresponding to ultraviolet light and red light, given that their wavelengths are 3000 Å and 7000 Å respectively. (Remember that photons are not coloured; colour is a human perception for that frequency range.)

(b) A typical FM radio station has its broadcast frequency 98.3 MHz. What is the energy of an FM photon of this frequency?

Given

\(\lambda_{uv} = 3000\) Å \(= 3000 \times 10^{-10}\) m,

\(\lambda_{red} = 7000\) Å \(= 7000 \times 10^{-10}\) m and

\(\nu_{FM} = 98.3\) MHz \(= 98.3 \times 10^{6}\ \mathrm{s^{-1}}\)

We know that the energy \(E\) of electromagnetic radiation of frequency \(\nu\) is \(h\nu\), and if \(\lambda\) is the corresponding wavelength, then \(\lambda\nu = c\), \(c\) being the speed of electromagnetic radiation in vacuum. The wavelength is given here rather than the frequency, so we substitute \(\nu = c/\lambda\) and use the combined form.

Hence, \(E = h\nu = \dfrac{hc}{\lambda}\)

\[E = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s} \times 3 \times 10^{8}\ \mathrm{m\,s^{-1}}}{3000 \times 10^{-10}\ \mathrm{m}} = 6.63 \times 10^{-19}\ \mathrm{J} = 4.144\ \mathrm{eV}\]

for a photon corresponding to ultraviolet light, and

\[E = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s} \times 3 \times 10^{8}\ \mathrm{m\,s^{-1}}}{7000 \times 10^{-10}\ \mathrm{m}} = 2.84 \times 10^{-19}\ \mathrm{J} = 1.776\ \mathrm{eV}\]

for a photon corresponding to red light.

  1. The energy of a photon of FM frequency 98.3 MHz is

\[E = 6.63 \times 10^{-34}\ \mathrm{J\,s} \times 98.3 \times 10^{6}\ \mathrm{s^{-1}} = 651.73 \times 10^{-28}\ \mathrm{J} = 40.74 \times 10^{-8}\ \mathrm{eV}\]

This is a very small energy as compared to the photon energy in the visible range.

Compare the three answers and the pattern of the whole chapter appears in miniature. The ultraviolet photon carries more than twice the energy of the red photon, and the FM photon carries less than a millionth of either. It is easy to see why ultraviolet radiation can throw electrons out of metals while an FM transmitter, radiating thousands of watts, cannot throw out a single one.

Two shortcuts that are useful while checking answers quickly: - Wavelength (in Å) × energy (in eV) ≈ 12500 (numerically) - Wavelength (in nm) × energy (in eV) ≈ 1250 (numerically)

Determine the wavelengths and frequencies for photons of energies (i) \(10^{-12}\) J, (ii) \(10^{-15}\) J, (iii) \(10^{-18}\) J, (iv) \(10^{-21}\) J and (v) \(10^{-24}\) J. Accordingly, prepare a chart along a horizontal line showing the various regions of the electromagnetic spectrum, and identify these regions in categories that you already know. Compare your results with a standard chart from any reference book or from the Internet. You would notice that γ photons are the most energetic photons, and their energies are about \(10^{-13}\) to \(10^{-12}\) J. This is a very small amount of energy on the human scale, and therefore we do not notice individual photons along their passage.

Explaining the Observations, One by One

The explanation using Einstein’s postulate of quantization of energy for the observations mentioned in Section 14.2.2 is given below. The numbering follows the same order, so each explanation can be read against the observation it accounts for.

Explanation 1. Einstein argued that when a photon of ultraviolet radiation arrives at the metal surface and collides with an electron, it gives all of its energy \(h\nu\) to the electron. The energy is gained by the electron and the photon no longer exists. There is no question of half a photon being absorbed; the transaction is all or nothing.

If \(\phi_0\) is the work function of the material of the emitter plate, then electrons will be emitted if and only if the energy gained by the electron is more than or equal to the work function, that is,

\[h\nu \ge \phi_0\]

Thus a minimum or threshold frequency \(\nu_0\) is required to eject electrons from the metal surface, where

\[\nu_0 = \frac{\phi_0}{h}\]

If \(\nu < \nu_0\), the photon will not have enough energy to liberate an electron. As a result, no electron will be ejected, however intense the incident radiation is. Similarly, if \(\nu > \nu_0\), the energy will always be sufficient to eject an electron, however small the incident intensity is.

NotePrinciple / Law

For photoelectric emission from a given metal of work function \(\phi_0\), the frequency \(\nu\) of the incident radiation must satisfy \(h\nu \ge \phi_0\); the threshold frequency is \(\nu_0 = \phi_0/h\), below which no emission occurs at any intensity.

Q1. Why does a metal give out electrons when light of one colour falls on it, but stays completely silent under another colour — however bright that light is made?

A. Because light delivers its energy one photon at a time, and each photon carries \(h\nu\), decided only by the colour (frequency). An electron must receive at least the work function \(\phi_0\) in a single transaction to escape. If the photons are too “small” — that is, if \(\nu < \nu_0\) — not one of them can lift an electron out, and sending in more of them by making the light brighter does not help, because the electron cannot add up two separate photons.

Imagine a wall that can only be crossed by jumping. If each of your jumps is too short, it makes no difference whether you jump once or a thousand times — you never get over. Making the light brighter is like jumping more often; raising the frequency is like jumping higher. Only the second one works.

Explanation 2. Energy is given by the photon to the electron as soon as the radiation is incident on the surface. The exchange of energy between the photon and the electron is instantaneous. Hence there is no time lag between the incidence of light and the emission of electrons. Also, when the incident radiation is stopped, there are no photons to transfer the energy to electrons, and hence the photoemission stops immediately.

Q2. When light strikes a metal, why do electrons come out at once, within a billionth of a second, instead of taking time to soak up enough energy?

A. Because an electron does not have to collect energy gradually. A single photon hands over its whole energy \(h\nu\) in one collision, and if that is at least equal to the work function, the electron leaves immediately. Waiting would only be necessary if energy arrived as a spread-out wave to be accumulated, which is exactly what the wave theory wrongly assumed.

Explanation 3. According to Einstein’s proposition, if the intensity of incident radiation for a given wavelength is increased, there will be an increase in the number of energy quanta (photons) incident on unit area in unit time, the energy of each quantum being the same, equal to \(h\nu = hc/\lambda\).

Therefore larger intensity radiation will knock off a larger number of electrons from the surface, and hence the current will be larger, provided \(\nu > \nu_0\). Conversely, lower intensity implies a smaller number of incident photons, hence a smaller number of ejected electrons and therefore a lower current.

In the photon picture, then, intensity is simply a photon count per second, not a measure of how energetic each photon is. This is the sentence to remember for every question that mixes up intensity and frequency.

Explanation 4. Once the electron is emitted from the surface, if the collector is at a higher potential than the emitter, the electric field will accelerate the electrons towards the collector. The higher the accelerating potential, the greater will be the number of electrons reaching the collector. Hence the photocurrent \(I\) increases with the accelerating potential initially.

Moreover, since the intensity of incident radiation determines the number of photons incident on the metal surface on unit area in unit time, it determines the maximum number of electrons that can be knocked off by the incident radiation. Hence, for a given intensity, increasing the accelerating potential can increase the current only till all the knocked-off electrons have reached the collector. No increase can be seen in the current beyond this limit. This explains the saturation current \(I_0\).

Explanation 5. Increasing the incident intensity will increase the number of incident photons and eventually the saturation current, since more electrons are now available to be collected.

Explanation 6. If the frequency of incident radiation is more than the threshold frequency, then the energy \(\phi_0\) is used by the electron to escape from the metal surface, and the remaining energy of the photon becomes the kinetic energy of the electron.

Depending on the energy of the electron inside the metal and other processes such as collisions after emission from the surface, the maximum kinetic energy is equal to \((h\nu - \phi_0)\). Hence

\[KE_{max} = h\nu - \phi_0 \qquad \text{--- (14.2)}\]

Equation (14.2) is known as Einstein’s photoelectric equation. Here \(KE_{max}\) is the maximum kinetic energy of the emitted photoelectrons in joule (J), \(h\nu\) is the energy of the incident photon in joule (J), and \(\phi_0\) is the work function of the emitter material in joule (J).

NotePrinciple / Law

Einstein’s photoelectric equation: for radiation of frequency \(\nu\) incident on a metal of work function \(\phi_0\), the maximum kinetic energy of the emitted photoelectrons is \(KE_{max} = h\nu - \phi_0\).

Read the equation as a simple statement of energy accounting, as shown in Fig. 14.2c: the photon brings in \(h\nu\), the metal charges a fixed toll \(\phi_0\) at the surface, and whatever is left over appears as kinetic energy of the escaping electron.

Diagram

Fig. 14.2c – Energy accounting in photoemission: an incoming photon of energy \(h\nu\) split into the work function \(\phi_0\) spent at the surface and the remainder appearing as \(KE_{max}\) of the electron.

\(KE_{max}\) depends on the material of the emitter plate and varies linearly with the incident frequency \(\nu\); it is independent of the intensity of the incident radiation.

Q. If every photon of a given beam carries the same energy \(h\nu\), why do the emitted electrons not all have the same kinetic energy?

A. Because the photon energy is identical, but the electrons are not. Each electron starts from a different situation inside the metal, and the equation quotes the best possible outcome. The reasons are listed in the next explanation.

Ultraviolet radiation of wavelength 2000 Å is incident on a zinc surface whose work function is 3.6 eV. Find (i) the energy of each incident photon in eV, (ii) the maximum kinetic energy of the emitted photoelectrons in eV, and (iii) the threshold wavelength for zinc.

Explanation 7. The electrons that are emitted from the metal surface have different kinetic energies. The reasons for this are manifold: all the electrons in a solid do not possess the same energy; the electrons may be ejected from varying depths inside the metal surface; and electrons may suffer collisions before they come out of the metal surface and may lose their energy.

If \(V\) is the potential difference between the emitter and the collector, and the collector is at a lower potential, an electron will lose its kinetic energy in overcoming the retarding force. If the kinetic energy is not sufficient, the emitted electrons may not reach the collector and the photocurrent will be zero.

If \(KE_{max}\) is the energy of the most energetic electron at the emitter surface, where its potential energy is zero, and \(-V_0\) is the stopping potential, then this electron will fail to reach the collector if \(KE_{max} < eV_0\), where \(e\) is the electron charge in coulomb (C) and \(eV_0\) is the energy needed for the electron to overcome the retarding potential \(V_0\).

If the electron just fails to reach the collector, that is, if it has lost all its kinetic energy exactly at the collector, then \(KE_{max} = eV_0\) and the photocurrent becomes zero. Equation (14.2) then explains that the stopping potential \(V_0\) depends on the incident frequency and on the material of the emitter, and does not depend on the incident intensity.

Explanation 8. If the ejected electrons have kinetic energy more than \(eV_0\), electrons can reach the collector and hence current flows. When the kinetic energy of the electron is less than or equal to \(eV_0\), no current will flow. The photocurrent will become zero when \(KE_{max} = eV_0\).

Using \(KE_{max} = eV_0\), we can write Eq. (14.2) as

\[eV_0 = h\nu - \phi_0\]

or,

\[V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad \text{--- (14.3)}\]

The above equation tells us that \(V_0\) varies linearly with the incident frequency \(\nu\), and the slope of the straight line depends on the constants \(h\) and \(e\), while the intercept of the line depends on the material through \(\phi_0\).

Compare this with the standard straight-line form \(y = mx + c\): here \(V_0\) is \(y\), \(\nu\) is \(x\), the slope \(m\) is \(h/e\), and the intercept \(c\) is \(-\phi_0/e\). Since \(h\) and \(e\) are universal constants, every metal must give a line of the same slope.

Thus the slopes of the lines in Fig. 14.6 are the same and are independent of the material of the emitter, but the intercepts are different for different materials. This is exactly what was observed, and it also gives experimenters a beautiful method of measuring \(h\): plot \(V_0\) against \(\nu\), measure the slope, and multiply by \(e\).

In a photoelectric experiment, light of frequency \(8.0 \times 10^{14}\) Hz falls on a sodium surface of work function 2.4 eV. Calculate the stopping potential. Taking \(h = 6.63 \times 10^{-34}\) J s and \(e = 1.6 \times 10^{-19}\) C, also find the threshold frequency of sodium.

Explanation 9. All the above arguments thus bring out the fact that the magnitude of the photocurrent depends on the incident intensity, through the number of emitted photoelectrons, and on the potential \(V\) of the collector, but not on the incident frequency \(\nu\) as long as \(\nu > \nu_0\).

How the Photon Idea Came to Be Accepted

Thus all the observations related to the experiments on photoelectric effect were explained by Einstein’s hypothesis of the existence of a photon, or treating light as bundles of energy.

Although Einstein gave his hypothesis in 1905, it was not widely accepted by the scientific community. The wave theory of light had a century of spectacular success behind it, and physicists were unwilling to give up any part of it on the strength of one idea.

In 1909, when Millikan measured the charge of an electron, and the value of \(h\) calculated from Eq. (14.3) matched with the value given by Planck, the hypothesis was accepted. The work function values \(\phi_0\) for some metals were also confirmed from Eq. (14.3). Einstein and Millikan received Nobel prizes for their respective discoveries in 1921 and 1923 respectively.

Millikan is a striking example of how science tests an idea it dislikes. He spent years on careful photoelectric measurements precisely because he expected Einstein’s equation to fail, and he said so openly. Instead, his data fitted Eq. (14.3) accurately and yielded a value of \(h\) in agreement with Planck’s. The verification was all the more convincing because it came from someone who had set out to disprove the theory.

You must have seen light emitting diodes (LEDs) of different colours. In an LED, electrical energy is converted into light energy corresponding to different colours. Can you tell what must be the difference in the working of LEDs of different colours? Design an experiment using LEDs to determine the value of Planck’s constant. You might know that the Nobel Prize in Physics for the year 2014 was awarded to Professors Isamu Akasaki, Hiroshi Amano and Shuji Nakamura for the invention of blue LEDs. They made the first blue LED in the early 1990s. Try to search on the Internet why it was difficult to make a blue LED.

Why We Never See Individual Photons

According to Einstein, the energy of radiation of frequency \(\nu\) comes in bundles with magnitude \(h\nu\). Thus the energy of a light beam having \(n\) photons will be \(nh\nu\), where \(n\) can take only integral values.

Is it then possible to vary the incident energy continuously? Why do we not see individual photons? To understand this issue, let us consider the following example.

The wavelength and power of the incident light are 4000 Å and 0.1 W respectively. What is the minimum change in the energy of the incident light? What is the number of incident photons?

Given incident intensity \(= 0.1\) W and \(\lambda = 4000\) Å \(= 4000 \times 10^{-10}\) m.

Since energy can change only by whole photons, the smallest possible change in the energy of the beam is the energy of a single photon. The energy \(E\) of a photon of the given wavelength is

\[E = h\nu = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s} \times 3 \times 10^{8}\ \mathrm{m/s}}{4000 \times 10^{-10}\ \mathrm{m}} = 4.972 \times 10^{-19}\ \mathrm{J}\]

This is the minimum change in energy, and it is very small. The change in energy can therefore be considered as continuous.

The number of photons \(N\) incident per second is obtained by dividing the energy supplied per second, that is the power, by the energy of one photon:

\[N = \frac{0.1\ \mathrm{W}}{4.97 \times 10^{-19}\ \mathrm{J}} \approx 2.011 \times 10^{17}\]

The number of photons coming out is so large that the human eye cannot comprehend or count it. Even if one wished to count, say, 10 photons per second, about \(10^{9}\) years would be required.

This is the resolution of an apparent paradox. Light really is granular, but the grains are so fine and so numerous that a beam looks perfectly smooth to us — much as a photograph looks continuous until you magnify it enough to see the individual dots.

A particular metal used as a cathode in an experiment on photoelectric effect does not show the photoelectric effect when it is illuminated with green light. Which of the colours in the visible spectrum are likely to generate photocurrent?

The Two Pictures Side by Side

Table 14.2 places each observation against what the wave theory predicts and what the photon picture predicts. Reading down the last two columns is the quickest way to revise this entire section.

Table 14.2: Summary of analysis of observations from experiments on photoelectric effect.

Observation Wave theory Photon picture
Electrons are emitted as soon as the light is incident on the metal surface. Very intense light is needed for instantaneous emission of electrons. Only one photon is needed to eject one electron from the metal surface, and energy exchange between electron and photon is instantaneous on collision.
Very low intensity of incident light is also sufficient to generate photocurrent. Low intensity should not give photocurrent. Low intensity of incident light means a smaller number of photons and not low energy photons. Hence a low current will be produced.
High intensity gives larger photocurrent, which means a higher rate of release of electrons. High intensity means higher energy radiation and therefore more electrons are emitted. Higher intensity means a larger number of photons incident in unit time, therefore a larger number of electrons are emitted in unit time and hence the photocurrent is larger.
Increasing the intensity has no effect on the electron energy. Higher intensity should mean electrons emitted with higher energies. Higher intensity means a higher number of incident photons per unit time. The energy of a photon is the same, as it does not depend on the intensity.
A minimum threshold frequency is needed for photocurrent to start. Low frequency light should release electrons but would take more time. A photon of low frequency light will not have sufficient energy to release an electron from the surface.
Increasing the frequency of incident light increases the maximum kinetic energy of electrons. Increasing intensity should increase the maximum kinetic energy. Maximum kinetic energy should not depend on the incident frequency. Increasing the frequency increases the energy of the photon. Therefore electrons receive more energy, which results in increasing the maximum kinetic energy.

In numerical problems from this section, almost everything reduces to three relations used together: \(E = h\nu = hc/\lambda\) for the photon, \(KE_{max} = h\nu - \phi_0\) for the emitted electron, and \(KE_{max} = eV_0\) for the stopping potential. The commonest errors are forgetting to convert eV to joule, and using wavelength in Å without converting to metre.

14.3 Wave-Particle Duality of Electromagnetic Radiation

From Energy Packets to Momentum

In its interaction with matter, light behaves as if it is made up of packets of energy called quanta. That much was established by the photoelectric effect.

Later it was confirmed from other theoretical and experimental investigations that these light quanta can have associated momentum as well. This is a significant step forward. Energy alone can be argued about; momentum is the property we normally reserve for something that has mass and moves — something we would unhesitatingly call a particle.

Hence the question came up whether a particle can be associated with light, or with electromagnetic radiation in general. The particle nature was confirmed by Compton in 1924, in experiments on the scattering of X-rays due to the electrons of matter.

The particle nature of radiation is seen in black body radiation and in the photoelectric effect. In the former, near room temperature, the radiation is mostly in the infrared region, while in the latter it is in the visible and ultraviolet region of the spectrum. The third experiment, which established that a photon possesses momentum like a particle, was Compton scattering, where X-rays and γ-rays interact with matter.

In 1923, A. H. Compton made a monochromatic beam of X-rays, of wavelength λ, incident on a graphite sheet and measured the intensity of the scattered rays in different directions as a function of wavelength. He found that although the incident beam consisted of a single wavelength λ, the scattered intensity was maximum at two wavelengths. One of these was the same as the incident wavelength, but the other, λ′, was larger by an amount Δλ. This Δλ is known as the Compton shift, and it depends on the scattering angle.

Compton explained his observations by considering the incidence of the X-ray beam on graphite as a collision of X-ray photons with the electrons of graphite, like a collision of billiard balls. Energy and momentum are transferred during the collision, and the scattered photons have lower energy than the incident photons. Therefore they have lower frequency, or higher wavelength.

The Compton shift is given by the relation

\[\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)\]

where θ is the scattering angle. The shift depends only on the scattering angle and not on the incident wavelength. This shift cannot be explained using wave theory. If we let Planck’s constant go to zero, we get the result expected from wave theory. This is the test to check whether the new picture is correct or not.

Diagram

Fig. 14.6d – Compton scattering: an X-ray photon of wavelength λ colliding with an electron in graphite, the scattered photon of longer wavelength λ′ leaving at angle θ and the recoiling electron moving away.

You can know more about these experiments from the reference books given at the end of this book, or from the links given below: - http://physics.usask.ca/~bzulkosk/modphyslab/phys251manual/compton_2009.pdf - http://www.phys.utk.edu/labs/modphys/Compton Scattering Experiment.pdf - http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html

Q. Why is a change in the wavelength of the scattered X-rays such strong evidence for the particle picture?

A. Because a wave scattered by matter should come back with the same frequency it went in with — the scatterer simply oscillates at the driving frequency and re-radiates at that frequency. A change of wavelength means the radiation actually lost a share of its energy in a collision, which is what particles do and waves do not. That is why the Compton shift settled the question.

The Properties of a Photon

Compton showed that a photon has an associated momentum along with the energy it carries. All photons of electromagnetic radiation of a particular frequency have the same energy and the same momentum.

Photons are electrically neutral, and so they are not deflected by electric or magnetic fields. This is one clear way in which a photon differs from an electron; a beam of electrons can be steered with a magnet, a beam of light cannot.

Photons can have particle-like collisions with other particles such as electrons. In a photon–particle collision, the energy and momentum of the system are conserved, but the number of photons is not conserved.

That last point deserves care, because it is unlike anything in mechanics. Photons can be absorbed, or new photons can be created. Photons can transfer their energy and momentum during collisions with particles and simply disappear. When we turn on a light, photons are created. There is no conservation law that fixes how many photons exist.

A photon always moves with the speed of light; it is never at rest. The mass of a photon is not defined as we do for a particle in Newtonian mechanics. Its rest mass is zero, in all frames of reference.

NoteDefinition

A photon is an electrically neutral quantum of electromagnetic radiation that carries both energy and momentum, always travels with the speed of light, has zero rest mass in all frames of reference, and can be created or destroyed in interactions.

Q. If a photon has zero rest mass, how can it carry momentum at all? In Newtonian mechanics, momentum is mass times velocity.

A. Because the Newtonian formula \(p = mv\) is not the general one. Einstein’s special theory of relativity provides a relation between energy and momentum that continues to work for a massless particle moving at the speed of light, and it assigns such a particle a definite non-zero momentum. The exact relation is written down in Section 14.5, where it is used to obtain the de Broglie relation.

Compton scattering is not merely a laboratory curiosity. In radiotherapy, the megavoltage X-ray beams used to treat tumours lose most of their energy in soft tissue precisely by Compton scattering off atomic electrons, and treatment planning software calculates doses using exactly this photon–electron collision picture. Radiation shielding for hospitals and reactors is designed on the same basis.

When Each Nature Shows Itself

Effects of the wave nature of light were seen in experiments on interference or diffraction when the slit widths, or the separation between two slits, are smaller than or comparable to the wavelength of light. If the slit width is large, or the spacing between slits is more, the interference or diffraction patterns will not be the same, and the wave nature will not be so obvious.

This gives a useful rule of thumb. The wave nature announces itself when the obstacle or opening is of about the same size as the wavelength; when the obstacle is enormous by comparison, light simply travels in straight lines and the wave character stays hidden.

It was realized by scientists that some phenomena observed in experiments in the laboratory or in nature, such as interference and diffraction, can be explained by considering light in particular, and electromagnetic radiation in general, as a wave.

On the other hand, some other observations, such as the photoelectric effect and black body radiation, can be explained only if we consider electromagnetic radiation as consisting of photons with a definite quantum of energy, and of momentum as well, as is evident from the Compton scattering experiments.

Also, there are some phenomena which can be explained by both the theories.

The Dual Character of Radiation

It is therefore essential to consider that both the characters or behaviours hold good; one dominates in some situations and the other works in the rest of the situations. It is necessary to keep both the physical models to explain the careful experimental observations. There is thus a need to hypothesize the dual character of light.

Note carefully what is being claimed and what is not. Light is not sometimes a wave and at other times secretly something else; nor does it change its nature depending on who is watching. It is a single entity whose behaviour requires two complementary descriptions, and which of the two descriptions is useful depends on the kind of experiment being done.

Later it turned out that such a picture is required not only for light but for the whole electromagnetic spectrum. This phenomenon is termed as wave-particle duality of electromagnetic radiation.

NoteDefinition

Wave-particle duality of electromagnetic radiation is the property by which electromagnetic radiation exhibits both wave-like behaviour, as in interference and diffraction, and particle-like behaviour, as in the photoelectric effect, black body radiation and Compton scattering, with one behaviour dominating in a given situation and the other in the rest.

In examinations, this idea is usually tested by asking you to place a given phenomenon in the correct column: interference, diffraction and polarization on the wave side; photoelectric effect, black body radiation and Compton scattering on the particle side. The reasoning to remember is that radiation travels as a wave but exchanges energy with matter as a particle.

14.4 Photo Cell

What a Photocell Is and How It Is Built

A photo cell is a device that makes use of the photoelectric effect and converts light energy into electrical energy. In other words, it is the photoelectric experiment of Section 14.2.1 turned into a practical component that can be wired into a circuit.

NoteDefinition

A photocell is a device that works on the photoelectric effect and converts light energy into electrical energy.

A schematic of a photocell is shown in Fig. 14.7. It consists of a semi-cylindrical photosensitive metal plate E, acting as a cathode, and a wire loop collector C, acting as an anode, both supported in an evacuated glass or quartz bulb.

Diagram

Fig. 14.7 – Schematic of a photocell: semi-cylindrical cathode E and wire loop anode C inside an evacuated bulb, connected through a high tension battery B and a microammeter, with incident light falling on the cathode.

The electrodes are connected to an external circuit having a high tension battery B and a microammeter μA. Instead of a photosensitive metal plate, the photosensitive material can be pasted in the form of a thin film on the inner walls of the glass bulb.

Each part of this design has a purpose. The cathode is made semi-cylindrical so that it presents a large curved surface to the incoming light, catching as much of the beam as possible. The anode is only a thin wire loop, so that it collects the electrons without standing in the way of the light. A microammeter is used because photocurrents are extremely small, of the order of microamperes.

How It Works

When light of suitable wavelength falls on the cathode, photoelectrons are emitted. These electrons are attracted towards the anode due to the applied electric field, and the generated photocurrent is noted from the microammeter.

The words “suitable wavelength” carry all the physics of the earlier sections. The material of the cathode must have a work function low enough that the light available in that situation crosses its threshold frequency. A photocell meant to work with visible light is therefore made with an alkali metal cathode, while a cell meant only for ultraviolet radiation can use a metal of higher work function.

Q. Why does a photocell need a battery at all, if light itself is supplying the energy that frees the electrons?

A. The light supplies the energy needed to pull electrons out of the cathode, but once they are out they would travel in all directions and many would never reach the thin anode wire. The battery sets up the accelerating field that sweeps them across to the anode, so that nearly every emitted electron is counted as current. The battery does not create the electrons; it only collects them.

A photocell in a light meter registers a photocurrent of \(4.8\ \mu\)A when a certain lamp is placed in front of it. Taking the charge on an electron as \(1.6 \times 10^{-19}\) C, calculate the number of photoelectrons reaching the anode per second. If the lamp is then replaced by one of twice the intensity but the same colour, what will the new reading be?

Where Photocells Are Used

A photocell is used to operate control systems and in light measuring devices. Light meters in photographic cameras make use of a photocell to measure the intensity of light. A photocell can also be used to switch on or off street lights.

Q4. How does a photocell switch street lights on at dusk and off at dawn, all by itself?

A. Daylight falling on the photosensitive cathode of the photocell releases photoelectrons, so a photocurrent flows as long as it is bright enough. This current holds a relay in the “lamp off” position. As evening comes, the intensity of daylight falls, the number of photons arriving each second drops, the photocurrent becomes too weak to hold the relay, and the circuit switches the street lamp on. At dawn the returning daylight restores the photocurrent and the lamp is switched off again. No person and no clock is involved — only the intensity of light.

Suppose a source of ultraviolet radiation is kept near the passage or entrance of a mall or a house, with the light made incident on the cathode of a photocell, so that a photocurrent is generated continuously. When a person passes through the passage, or comes near the entrance, the incident light beam is interrupted and the photocurrent stops. This event can be used to operate a counter in counting devices, or to set a burglar alarm. Such an arrangement can also be used to identify traffic law defaulters by setting off an alarm using the photocell.

The same idea, drawn in Fig. 14.4a, lies behind the automatic doors at shop entrances and the safety beam across a lift door that keeps it from closing on a passenger.

Diagram

Fig. 14.4a – A light source and a photocell facing each other across a passage; a person walking through interrupts the beam and the photocurrent falls to zero, triggering a counter or alarm.

Notice how directly this application uses Observation 3 of Section 14.2.2. The photocurrent depends on the intensity of the light reaching the cathode, so anything that blocks the light produces an instant electrical signal. Because photoemission stops the moment the radiation stops, the response is effectively immediate — which is exactly what a counter or an alarm needs.

Is a solar cell a photocell?

14.5 De Broglie Hypothesis

An Argument from Symmetry

In 1924, Prince Louis de Broglie (pronounced as “de broy”) proposed, on the basis of the symmetry existing in nature, that if radiation has a dual nature — sometimes wave nature dominating and sometimes particle nature — then matter may also possess a dual nature.

The reasoning deserves a moment’s thought, because it is unusually daring. There was, at that time, no experiment at all suggesting that matter behaves like a wave. De Broglie’s argument was that nature does not usually treat its two great categories, radiation and matter, by two different rules. If light, long believed to be purely a wave, had turned out to be partly a particle, then matter, long believed to be purely particles, might well turn out to be partly a wave.

Normally we talk about matter as being composed of particles, but are there situations where matter seems to show wave-like properties? This will become evident from the experiments on diffraction of electrons from nickel crystals, described later in this chapter.

Building the Relation

De Broglie used the properties frequency \(\nu\) and wavelength \(\lambda\) of a wave and proposed a relation to connect these with the particle properties, energy \(E\) and momentum \(p\).

He started from what was already known about the photon. The momentum \(p\) carried by a photon of energy \(E\) is given by the relation

\[p = \frac{E}{c} \qquad \text{--- (14.4)}\]

which is valid for a massless particle travelling with the speed of light \(c\), according to Einstein’s special theory of relativity. Here \(p\) is the momentum in kg m s\(^{-1}\), \(E\) is the energy in joule (J) and \(c\) is the speed of light in vacuum in m s\(^{-1}\). This is the promised relation that gives a photon a finite momentum despite its zero rest mass.

Using Einstein’s relation for \(E\), that is \(E = h\nu\),

\[p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda} \qquad \text{--- (14.5)}\]

where \(\lambda\), the wavelength, is given by \(\lambda\nu = c\).

Look carefully at the last step in Eq. (14.5). It links the momentum of the photon — a particle property — directly to its wavelength, a wave property, through Planck’s constant. This is the equation de Broglie chose to carry over to matter.

De Broglie proposed that a moving material particle of total energy \(E\) and momentum \(p\) has associated with it a wave analogous to a photon. He then suggested that the wave and particle properties of matter can also be described by a relation similar to Eq. (14.5) for a photon.

Thus the frequency and wavelength of a wave associated with a material particle, of mass \(m\) moving with a velocity \(v\), are given as

\[\nu = \frac{E}{h} \quad \text{and} \quad \lambda = \frac{h}{p} = \frac{h}{mv} \qquad \text{--- (14.6)}\]

Here \(m\) is the mass of the particle in kilogram (kg), \(v\) is its velocity in m s\(^{-1}\), \(p = mv\) is its momentum in kg m s\(^{-1}\), and \(\lambda\) is the associated wavelength in metre (m).

He referred to these waves associated with material particles as matter waves. The wavelength of the matter waves, given by Eq. (14.6), is now known as the de Broglie wavelength. Greater the momentum, shorter is the wavelength. Equation (14.6) for the wavelength of matter waves is known as the de Broglie relation.

NotePrinciple / Law

De Broglie hypothesis: every moving material particle has a wave associated with it, whose wavelength is \(\lambda = h/p = h/mv\), where \(p\) is the momentum of the particle. Such waves are called matter waves.

NoteDefinition

The de Broglie wavelength of a moving particle is the wavelength \(\lambda = h/p\) of the matter wave associated with it, where \(p\) is the momentum of the particle.

A moving particle is therefore to be pictured as in Fig. 14.5a — not as a hard little ball, but as a ball accompanied by a wave whose wavelength shrinks as the particle is made faster or heavier.

Diagram

Fig. 14.5a – A moving particle with its associated matter wave, drawn twice: a slow particle with a long wavelength and a fast particle with a short wavelength.

Q5. If light can behave like a stream of particles, can an electron behave like a wave?

A. Yes. De Broglie proposed in 1924 that every moving particle has a wave associated with it, of wavelength \(\lambda = h/mv\). For an electron, this wavelength works out to a few ångström — small, but not impossibly small. Within three years this prediction was confirmed experimentally: electrons fired at a crystal produced a diffraction pattern with maxima and minima, something only waves can do.

Q. If an electron has a wave associated with it, what exactly is waving?

A. Nothing material is waving — there is no substance moving up and down, as there is in a water wave or a stretched string. The matter wave is not a disturbance in a medium; it is a wave that tells us where the particle is likely to be found. What can be said with certainty at this level is that this wave shows the behaviour that defines a wave — interference and diffraction — and that its wavelength is exactly \(h/p\). The deeper interpretation belongs to quantum mechanics, which you will meet in higher classes.

Useful Forms of the de Broglie Relation

In problems, the speed of the particle is usually not given directly. What is given is its kinetic energy, or the potential difference through which it has been accelerated. So it is worth rewriting Eq. (14.6) in those terms.

For a particle of mass \(m\) moving with a velocity \(v\), the kinetic energy is

\[E_K = \frac{1}{2}mv^2 \quad \text{or} \quad v = \sqrt{\frac{2E_K}{m}}\]

Thus,

\[\lambda = \frac{h}{mv} = \frac{h}{m}\sqrt{\frac{m}{2E_K}} = \frac{h}{\sqrt{2mE_K}}\]

For a charged particle of charge \(q\), accelerated from rest through a potential difference \(V\), the work done is \(qV\). This work provides the kinetic energy of the particle. Thus \(E_K = qV\), and therefore

\[\lambda = \frac{h}{\sqrt{2mE_K}} = \frac{h}{\sqrt{2mqV}}\]

Here \(q\) is the charge in coulomb (C) and \(V\) is the accelerating potential difference in volt (V).

This relation holds for any charged particle, like an electron, a proton, or even a charged ion, where \(m\) corresponds to the mass of the charged particle. Of course, when \(V\) is very large — say in kilovolts — so that the speed of the particle becomes close to the speed of light, such an equation will not be applicable. You will learn about the other effects that arise in such situations in higher classes.

Q. Among an electron, a proton and an α-particle accelerated through the same potential difference, which will have the longest de Broglie wavelength?

A. Use \(\lambda = h/\sqrt{2mqV}\). With \(V\) common to all three, the wavelength is decided by the product \(mq\) — the larger this product, the shorter the wavelength. The electron has by far the smallest mass and the same magnitude of charge as the proton, so the electron has the longest wavelength. This one line of reasoning settles a whole family of examination questions.

The Working Formula for an Electron

For an electron moving through a potential difference of \(V\) (given in volts), substituting the mass and charge of the electron gives

\[\lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s}}{\sqrt{2 \times 9.11 \times 10^{-31}\ \mathrm{kg} \times 1.6 \times 10^{-19}\ \mathrm{C} \times V\ (\text{in volts})}} = \frac{1.228 \times 10^{-9}}{\sqrt{V\ (\text{in volts})}}\ \mathrm{m}\]

or,

\[\lambda\ (\text{in nm}) = \frac{1.228}{\sqrt{V\ (\text{in volts})}} \qquad \text{--- (14.7)}\]

Equation (14.7) is worth memorizing, but only along with its conditions: it applies to an electron, accelerated from rest, through a potential difference \(V\) expressed in volts, and it gives the answer directly in nanometre.

An electron is accelerated through a potential of 120 V. Find its de Broglie wavelength.

Given \(V = 120\) V.

We know that \(\lambda = \dfrac{1.228}{\sqrt{V}}\) using Eq. (14.7), with \(\lambda\) in nanometre when \(V\) is in volt.

\[\therefore\ \lambda = \frac{1.228}{\sqrt{120}} = 0.112\ \mathrm{nm}\]

An answer of about a tenth of a nanometre, that is about 1 Å, is worth remembering as a rough guide. It is the size of an atom — and that fact is the key to the whole of the next section.

A proton and an electron are each accelerated from rest through a potential difference of 100 V. Taking the mass of the proton as 1836 times that of the electron, find the ratio of their de Broglie wavelengths, and calculate the de Broglie wavelength of the electron in nanometre.

Can you estimate the de Broglie wavelength of the Earth?

The expression \(p = E/c\) defines the momentum of a photon. Can this expression be used for the momentum of an electron or a proton?

The Prediction Is Confirmed

Shortly after the existence of photons — particles associated with electromagnetic waves — was postulated, it was also experimentally found that sub-atomic particles like electrons, protons and neutrons, and atomic particles, also exhibit wave properties.

The wavelength associated with an electron of energy a few eV is of the order of a few Å. Therefore, to observe the wave nature of an electron, the slit width or the diffracting objects should be of the same order of magnitude, that is, a few Å.

This is the same condition that appeared at the end of Section 14.3 for light. No ordinary slit cut with a blade is anywhere near a few ångström wide, so an entirely different kind of “slit” had to be found.

The wave property of the electron was confirmed experimentally in 1927 by Davisson and Germer in America, and in 1928 by George P. Thomson in England, by diffraction of electrons by atoms in metals.

Knowing that the size of atoms and their spacing in crystals is of the order of a few Å, they anticipated that if electrons are scattered by atoms in a crystal, the associated matter waves will interfere and will show diffraction effects. A crystal, in other words, is a ready-made diffraction grating of exactly the right spacing, supplied by nature.

It turned out to be true in their experiments. Electrons showed constructive and destructive interference. No electrons were found in certain directions, due to destructive interference, while in other directions maximum numbers of electrons were seen, due to constructive interference.

Louis de Broglie received the Nobel Prize in Physics in 1929, and Davisson, Germer and Thomson shared the Nobel Prize in Physics in 1937. There is a remarkable family story hidden in these awards: Sir J. J. Thomson discovered the existence of the electron as a sub-atomic particle, while his son G. P. Thomson showed that the electron behaves like a wave. Father and son were each right, and each received a Nobel Prize — which is perhaps the neatest illustration of wave-particle duality that the history of physics offers.

14.6 Davisson and Germer Experiment

The Arrangement

A schematic of the experimental arrangement of the Davisson and Germer experiment is shown in Fig. 14.8. The whole set-up is enclosed in an evacuated chamber, for the same reason as in the photoelectric apparatus: a stray gas molecule in the path would scatter the electrons and destroy the pattern being looked for.

Diagram

Fig. 14.8 – Schematic of Davisson and Germer experiment: electron gun with filament F and battery B, cylindrical anode producing a collimated electron beam, nickel crystal target, movable electron detector on a circular scale, and galvanometer.

The set-up uses an electron gun — a device to produce electrons by heating a tungsten filament F using a battery B. A hot filament releases electrons from its surface, which is the simplest way of obtaining a supply of free electrons.

Electrons from the gun are accelerated through vacuum to a desired velocity by applying a suitable accelerating potential across a cylindrical anode, and are collimated into a focused beam. The accelerating potential is the crucial control here: as Eq. (14.7) shows, fixing the voltage fixes the de Broglie wavelength of the electrons in the beam.

This beam of electrons falls on a nickel crystal and is scattered in different directions by the atoms of the crystal. Thus, in the Davisson and Germer experiment, electrons were used in place of light waves, and a crystal was used in place of a diffraction grating.

Scattered electrons were detected by an electron detector, and the current was measured with the help of a galvanometer. The current measured is a direct count of how many electrons are arriving in that particular direction each second.

By moving the detector on a circular scale — that is, by changing the scattering angle \(\theta\) — the intensity of the scattered electron beam was measured for different values of the scattering angle.

NoteDefinition

The scattering angle \(\theta\) is the angle between the direction of the incident electron beam and the direction in which the scattered electrons are detected.

Q. Why use a crystal instead of cutting a very narrow slit?

A. Because the de Broglie wavelength of an electron accelerated through a few tens of volts is only about 1 Å, and no slit can be cut that narrow. In a crystal, however, the atoms are already arranged in regular layers spaced a few ångström apart. Nature has supplied a grating of exactly the right spacing, and all the experimenter has to do is aim the beam at it.

What Was Observed

The scattered intensity was not found to be uniform in all directions, as had been predicted by classical theory. If electrons were simply small bullets bouncing off atoms, the scattered electrons would be spread out smoothly, with no direction sharply favoured over its neighbours.

Instead, the intensity pattern resembled a diffraction pattern, with peaks corresponding to constructive interference and troughs corresponding to regions of destructive interference.

Diffraction is a property of waves. Hence the above observations implied that the electrons formed a diffraction pattern on scattering, and that particles could show wave-like properties.

The experiment owes a good deal to an accident. In the mid-1920s, while Davisson and Germer were studying the scattering of electrons from a nickel target, the vacuum in their apparatus was broken and air reached the hot target, oxidizing its surface. To clean it, they heated the nickel strongly for a long time. That heating recrystallized the metal: instead of a jumble of tiny grains, the target now consisted of a few large crystals with atoms in regular rows. When the scattering measurements were repeated, sharp peaks appeared where none had been seen before — and those peaks turned out to be the first clear evidence of electron diffraction.

The Numerical Test

Davisson and Germer varied the accelerating potential from 44 V to 68 V and observed a peak in the intensity of the scattered electrons at a scattering angle of 50º for a potential of 54 V.

This peak was the result of constructive interference of the electrons scattered from different layers of the regularly spaced atoms of the nickel crystal — waves reflected from successive atomic layers arriving in step with one another, exactly as in the diffraction of light from a grating.

The real test, however, is quantitative. The de Broglie relation predicts a definite wavelength for 54 V electrons, and the diffraction pattern independently measures a wavelength. If the two numbers agree, the hypothesis stands.

From Eq. (14.7), for \(V = 54\) V, we get

\[\lambda = \frac{1.228}{\sqrt{54}} = 0.167\ \mathrm{nm} \qquad \text{--- (14.8)}\]

From the electron diffraction measurements, the wavelength of the matter waves associated with the electrons was found to be 0.165 nm.

The two values of \(\lambda\), one obtained from the experimental results and the other from the theoretical de Broglie relation, were in close agreement. The Davisson and Germer experiment thus substantiated de Broglie’s hypothesis of wave-particle duality and verified his relation.

Q. The two values, 0.167 nm and 0.165 nm, are not identical. Does that weaken the conclusion?

A. No. A difference of about one part in a hundred is well within the uncertainties of such measurements, and what matters is that theory and experiment agree on the same value to three digits. Remember also that de Broglie’s relation was not fitted to this data — it was written down years earlier from an argument about symmetry, with no experiment to guide it. Agreement of that kind is strong evidence.

In an electron diffraction experiment, electrons are accelerated from rest through a potential difference of 100 V. Calculate their de Broglie wavelength in nanometre using Eq. (14.7), and compare it with the typical spacing between atomic layers in a crystal, about 0.2 nm. State whether a clear diffraction pattern is to be expected in this case.

The diffraction results described above can be produced in the laboratory using an electron diffraction tube. It has a filament which, on heating, produces electrons; this filament acts as a cathode. The electrons are accelerated to quite high speeds by creating a large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on a phosphor screen. By changing the voltage between the cathode and the anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why.

Diagram

Fig. 14.6a – Electron diffraction tube: hot cathode and electron beam, thin graphite target, and diffraction rings of radius \(r\) formed on the phosphor screen at a distance L from the target.

In examinations, this section is most often asked as “state the importance of the Davisson and Germer experiment”. The expected answer has two parts: it demonstrated the diffraction of electrons, thereby establishing the wave nature of matter, and the wavelength it measured agreed with the value calculated from the de Broglie relation, thereby verifying that relation quantitatively.

14.7 Wave-Particle Duality of Matter

Two Vocabularies for One Reality

Material particles show wave-like nature under certain circumstances. This phenomenon is known as wave-particle duality of matter.

NoteDefinition

Wave-particle duality of matter means that material particles possess a wave character in addition to their particle character, and show wave-like behaviour under suitable circumstances.

Frequency \(\omega\) and wave number \(k\) are used to describe waves in classical theories, while mass \(m\) and momentum \(p\) are used to describe particles. These are two separate sets of words, developed for two kinds of objects that were believed to have nothing in common.

Wave-particle duality implies that all moving particles have an associated frequency and an associated wave number, and all waves have an associated energy and an associated momentum. The two vocabularies, in other words, describe the same objects, and the de Broglie relation is the dictionary that translates between them.

Q. What is the wave number of a wave, and how is it related to the wavelength?

A. The wave number tells you how many wave cycles are packed into unit length, so it is inversely related to the wavelength — a short wavelength means a large wave number. Since a particle’s wavelength is \(h/p\), a particle with large momentum corresponds to a wave with a large wave number.

We come across the wave-particle duality of matter, due to quantum behaviour, when we are dealing with microscopic objects, of sizes \(\le 10^{-6}\) m. The small order of magnitude of \(h\) sets the scale at which quantum phenomena manifest themselves.

That last sentence explains a great deal. Planck’s constant is about \(6.6 \times 10^{-34}\) J s, an extraordinarily small number, and it sits in the numerator of \(\lambda = h/p\). Any object with an everyday momentum therefore gets an absurdly small wavelength. If \(h\) had been a large number, the world around us would look thoroughly quantum.

Why We Never Notice the Wave Nature of Everyday Objects

If all the material objects in motion have an associated wavelength, and therefore an associated wave, why then do we not talk about the wavelength of a child running with speed \(v\) on a pathway 2 m wide, or of a car moving with speed \(v\) on a road 20 m wide? To understand this, let us try to calculate these quantities.

A student, weighing 45 kg, is running with a speed of 8 km per hr on a foot path 2 m wide. A small car, weighing 1200 kg, is moving with a speed of 60 km per hr on a 20 m wide road. Calculate their de Broglie wavelengths.

Given

\(v_1 = 8\) km/hr \(= 8 \times 10^{3}/3600\) m/s and \(m_1 = 45\) kg for the student,

\(v_2 = 60\) km/hr \(= 60 \times 10^{3}/3600\) m/s and \(m_2 = 1200\) kg for the car.

Since \(\lambda = h/p\), the first step in each case is to find the momentum \(p = mv\).

momentum \(p_1 = 45 \times 8 \times 10^{3}/3600 = 100\) kg m/s for the student, and

momentum \(p_2 = 1200 \times 60 \times 10^{3}/3600 = 20000\) kg m/s for the car.

The de Broglie wavelength

\[\lambda_1 = \frac{h}{p_1} = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s}}{100\ \mathrm{kg\,m/s}} = 6.63 \times 10^{-36}\ \mathrm{m}\]

for the student, and the de Broglie wavelength

\[\lambda_2 = \frac{h}{p_2} = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s}}{20000\ \mathrm{kg\,m/s}} = 3.32 \times 10^{-38}\ \mathrm{m}\]

for the car.

The wavelengths calculated in Example 14.5 are negligible compared to the size of the moving objects, as well as to the widths of the paths on which the objects are moving.

To feel how negligible, compare \(10^{-36}\) m with the size of an atomic nucleus, about \(10^{-15}\) m. The student’s wavelength is smaller than a nucleus by a factor of a billion billion. There is no obstacle anywhere in the universe narrow enough to diffract it.

Diagram

Fig. 14.7a – A scale line marked with the width of a footpath (metres), the size of an atom (\(10^{-10}\) m), the size of a nucleus (\(10^{-15}\) m) and the de Broglie wavelength of a running student (\(10^{-36}\) m), showing how far off the scale the last one lies.

Therefore the wavelengths associated with macroscopic particles do not play any significant role in our everyday life, and we need not consider their wave nature. Also, the wavelengths for macroscopic particles are so small that they cannot be measured.

Q6. Every moving object is said to have a wavelength. Then why do we never notice the wave nature of a running student or a car on a highway?

A. Because the wavelength is \(\lambda = h/mv\), and Planck’s constant \(h\) is fantastically small while the momentum of an everyday object is large. A running student of 45 kg has a de Broglie wavelength of about \(6.63 \times 10^{-36}\) m, and a moving car about \(3.32 \times 10^{-38}\) m. Wave effects such as diffraction show up only when the wavelength is comparable to the size of the opening or obstacle, and nothing that small exists. The wave nature is not absent — it is simply unobservable, and far too small to measure.

On the other hand, if we try to estimate the associated de Broglie wavelength of a moving electron passing through a small aperture of size \(10^{-10}\) m, or of an oxygen molecule in air, we will find it to be significant, as can be seen in the following example.

Calculate the de Broglie wavelength of an electron moving with kinetic energy of 100 eV, passing through a circular hole of diameter 2 Å.

Given

\(E_K = 100\) eV \(= 100 \times 1.6 \times 10^{-19}\) J.

Since the kinetic energy is given, we first obtain the speed of the electron, then its momentum, and only then the wavelength. The speed of the electron is given by the relation

\[\frac{1}{2}mv^2 = 100 \times 1.6 \times 10^{-19}\ \mathrm{J}\]

\[\therefore\ v = \sqrt{\frac{2 \times 100 \times 1.6 \times 10^{-19}\ \mathrm{J}}{9.11 \times 10^{-31}\ \mathrm{kg}}} = 5.927 \times 10^{7}\ \mathrm{m/s}\]

and

momentum \(p = 9.11 \times 10^{-31}\) kg \(\times\ 0.593 \times 10^{7}\) m/s \(= 5.40 \times 10^{-24}\) kg m/s

\(\therefore\) the de Broglie wavelength

\[\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\ \mathrm{J\,s}}{5.40 \times 10^{-24}\ \mathrm{kg\,m/s}} = 1.23 \times 10^{-10}\ \mathrm{m} = 1.228\ \text{Å}\]

The wavelength of the electron in the above example is comparable to the size of the hole through which the electron is passing. The wavelength associated with this electron is the same as the size of a helium atom, and more than double the size of a hydrogen atom.

That comparison is the whole point. For the running student, the wavelength was smaller than the path by a factor of \(10^{36}\); for this electron, the wavelength and the hole are of the same size. Under such conditions the electron cannot be treated as a tiny bullet — it will diffract on its way through.

On what scale, or under which circumstances, is the wave nature of matter apparent?

When Newtonian Mechanics Is Still Enough

The photon picture allows transfer of energy and momentum in the same manner as in Newtonian mechanics. The wave nature does not modify that.

Whenever the wavelengths are small compared to the dimensions of slits or obstacles, or are not measurable, we can use Newtonian mechanics. This is a reassuring statement: quantum physics does not throw away the mechanics you have learnt, it restricts it to the domain where it works, which happens to include the whole of ordinary engineering and daily life.

In conclusion, for both electromagnetic radiation and atomic and sub-atomic particles, the particle nature is dominant during their interaction with matter. On the other hand, while travelling through space, particularly when their confinement is of the same order of magnitude as their associated wavelength, the wave nature is dominant.

Q. How do I decide, in an examination question, whether to treat something as a wave or as a particle?

A. Ask two questions. First, is the object interacting with matter — being absorbed, emitted, or colliding? Then use the particle picture. Second, is it travelling through, or confined by, something of about its own wavelength — a slit, a crystal, an aperture? Then use the wave picture. Applying both tests to the photoelectric effect gives “particle”, and to electron diffraction gives “wave”.

We have seen earlier that electrons are bound inside a metal surface and need some minimum energy, equal to the work function, to be knocked off from the surface. This energy, if provided by any means, can make the electron come out of the metal surface. The physical ways used to provide this energy differentiate the physical processes involved, and accordingly different devices and characterizing microscopes based on them have been designed by scientists. - Thermionic emission: by heating to temperatures of about 2000 ºC, thermal energy is provided. - Field emission: by establishing strong electric fields of about \(10^{6}\) V/m at the surface of a metal tip, electrical energy is provided. - Photo-electron emission: by shining radiation of suitable frequency, ultraviolet or visible, on a metal surface, light energy is provided.

Seeing with Electrons

You have learnt about the resolving power and the resolution of telescopes and microscopes that use ordinary visible light. The resolution of a microscope is limited by the wavelength of the light used.

The shorter the wavelength of the characterizing probe, the smaller is the limit of resolution of a microscope, that is, the better is the resolution of the microscope. Better resolution can therefore be attained by illuminating the objects to be seen by radiation of smaller wavelengths. No microscope can show clear detail much finer than the wavelength it uses to look, which is why an optical microscope cannot resolve objects much smaller than a few hundred nanometre.

We have seen that an electron can behave as a wave, and its wavelength is much smaller than the wavelength of visible light. The wavelength can be made much smaller still, as it depends on the velocity and the kinetic energy of the electron.

An electron beam accelerated to several keV of energy will correspond to a de Broglie wavelength much smaller than an angstrom, that is, \(\lambda_e \ll 1 \times 10^{-10}\) m. The resolution of such an electron microscope will be several hundred times higher than that obtainable with an optical microscope.

Other advantages of electron microscopes are that (i) electrons do not penetrate matter as visible light or X-rays do, (ii) electron beams can be more easily produced and controlled by electric and magnetic fields than electromagnetic waves, and (iii) electrons can be focused just as light is focused with lenses.

It was proposed in 1925 that atoms in solids can act as diffraction centres for electron waves and can give information about the geometry or structure of a solid, just as X-rays do on getting diffracted by solids. However, it took many years to realize an electron microscope for practical applications. The first electron microscope was developed by Ruska, in Berlin, Germany, in the year 1929.

Microscopic objects, when illuminated using electron beams, yield high resolution images. Images of microscopic and nanometric objects, and even of viruses, have been obtained by scientists using electron microscopes, making valuable contributions to mankind. Transmission electron microscopy can resolve very small particles: the micrograph on the cover page of this book shows tiny crystals of dimensions less than 50 nm. An electron diffraction pattern is also seen on the cover page, as a spot pattern. When an electron beam passes through a crystal having a periodic arrangement of atoms, diffraction occurs, because the crystal acts as a collection of diffraction slits for the electron beam.

Notice that the electron microscope is the practical pay-off of this entire chapter. It works only because electrons have a wavelength, only because that wavelength can be shortened by accelerating them, and only because a crystal can diffract them — three results established in Sections 14.5 and 14.6.

Internet my friend — for further reading on the topics of this chapter: 1. http://phet-web.colorado.edu/simulations/schrodinger/dg.jnlp 2. https://physics.info/photoelectric/ 3. https://www.britannica.com/science/photoelectric-effect 4. https://www.britannica.com/science/wave-particle-duality 5. https://www.sciencedaily.com/terms/wave-particle_duality.htm 6. https://www.thoughtco.com/de-broglie-hypothesis-2699351 7. https://www.toppr.com/guides/physics/dual-nature-of-radiation-and-matter 8. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/DavGer2.html

14.8 Summary

Quantization of Energy
  1. Electromagnetic radiation consists of mutually perpendicular oscillating electric and magnetic fields, both perpendicular to the direction in which the wave and its energy travel. Interference, diffraction and polarization are explained by this wave picture.
  2. The black body radiation spectrum could not be explained by the wave nature of radiation. Planck explained it by proposing that energy is emitted in packets, and that at higher frequencies the energy of a packet is large.
  3. Planck assumed that atoms behave like tiny oscillators that emit electromagnetic radiation only in discrete packets, \(E = nh\nu\), where \(\nu\) is the frequency of the oscillator and \(n\) is an integer. Emission occurs only when the oscillator jumps from one quantized energy level to another of lower energy.
The Photoelectric Effect
  1. The photoelectric effect is the emission of electrons from a metal surface when radiation of appropriate frequency is incident on it. A surface that emits electrons on illumination is called a photosensitive surface, and the emitted electrons are called photoelectrons.
  2. Ultraviolet radiation is necessary for metals like zinc, cadmium and magnesium, while for alkali metals even visible radiation is sufficient.
  3. The work function \(\phi_0\) of a metal is the minimum energy required to pull an electron out of the metal from its surface. It depends on the metal and on the nature of its surface, and is usually expressed in electron volt (eV).
  4. Key experimental observations: emission occurs only if \(\nu > \nu_0\), the threshold frequency, whatever the intensity; emission begins within \(10^{-9}\) s, with no time lag; photocurrent increases linearly with intensity; photocurrent saturates at \(I_0\) on increasing the accelerating potential; the saturation current increases proportionately with intensity; \(KE_{max}\) depends on the frequency and the emitter material but not on the intensity; the photocurrent falls to zero at the stopping potential \(-V_0\); and \(V_0\) varies linearly with \(\nu\), the lines for different metals having the same slope but different intercepts.
Failure of the Wave Theory
  1. The wave theory predicts that energy is absorbed continuously and shared among all surface electrons, so a waiting period of minutes to hours would be needed before emission — of the order of \(4.428 \times 10^{4}\) s, about half a day, for the case worked out in Example 14.1. Emission is in fact instantaneous.
  2. The wave theory predicts that higher intensity should give more energetic electrons and that any frequency should work if the light is bright enough. Both predictions are contradicted by experiment.
Einstein’s Photon Picture
  1. Einstein extended Planck’s quantization to all electromagnetic radiation in 1905, proposing that light energy is released or absorbed in bundles called photons, of energy \[E = h\nu \qquad \text{--- (14.1)}\] where \(h = 6.626 \times 10^{-34}\) J s is Planck’s constant. Since \(\nu = c/\lambda\), energy is directly proportional to frequency and inversely proportional to wavelength: \(E = hc/\lambda\).
  2. A photon gives its whole energy \(h\nu\) to a single electron in one instantaneous collision, and then ceases to exist. Emission occurs if and only if \(h\nu \ge \phi_0\), so the threshold frequency is \(\nu_0 = \phi_0/h\).
  3. Intensity fixes the number of photons arriving per unit area per unit time, and therefore the number of photoelectrons and the photocurrent. Frequency fixes the energy of each photon, and therefore the maximum kinetic energy of the photoelectrons.
  4. Einstein’s photoelectric equation: \[KE_{max} = h\nu - \phi_0 \qquad \text{--- (14.2)}\]
  5. At the stopping potential the most energetic electron just fails to reach the collector, so \(KE_{max} = eV_0\), which gives \[V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e} \qquad \text{--- (14.3)}\] The slope \(h/e\) is the same for all metals; the intercept depends on \(\phi_0\) and hence on the material.
  6. Millikan’s measurements in 1909 confirmed the value of \(h\) obtained from Eq. (14.3), after which the photon hypothesis was accepted. Einstein and Millikan received Nobel Prizes in 1921 and 1923 respectively.
  7. Useful numerical checks: wavelength (in Å) × energy (in eV) ≈ 12500, and wavelength (in nm) × energy (in eV) ≈ 1250.
Wave-Particle Duality of Radiation
  1. Compton confirmed the particle nature of radiation through the scattering of X-rays by electrons. The scattered radiation contains a longer wavelength \(\lambda'\), and the Compton shift is \[\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)\] where \(\theta\) is the scattering angle. The shift depends only on \(\theta\) and not on the incident wavelength, and cannot be explained by wave theory.
  2. A photon carries momentum as well as energy, is electrically neutral, is not deflected by electric or magnetic fields, always moves with the speed of light, and has zero rest mass in all frames of reference. In photon–particle collisions, energy and momentum are conserved but the number of photons is not.
  3. Wave behaviour of light shows up when slit widths or slit separations are smaller than or comparable to the wavelength. Since some phenomena need the wave model and others need the photon model, both must be retained; one dominates in some situations and the other in the rest. This is the wave-particle duality of electromagnetic radiation.
Photocell
  1. A photocell uses the photoelectric effect to convert light energy into electrical energy. It has a semi-cylindrical photosensitive cathode and a wire loop anode in an evacuated glass or quartz bulb, connected to a high tension battery and a microammeter.
  2. Photocells are used in control systems, in light meters of photographic cameras, for switching street lights on and off, and — by interrupting a light beam — in counters and burglar alarms.
De Broglie Hypothesis
  1. For a photon, momentum and wavelength are related by \[p = \frac{E}{c} \qquad \text{--- (14.4)}\] \[p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda} \qquad \text{--- (14.5)}\]
  2. De Broglie proposed in 1924 that a moving material particle also has a wave associated with it, with \[\nu = \frac{E}{h} \quad \text{and} \quad \lambda = \frac{h}{p} = \frac{h}{mv} \qquad \text{--- (14.6)}\] These are called matter waves, and \(\lambda\) is the de Broglie wavelength. The greater the momentum, the shorter the wavelength.
  3. In terms of kinetic energy, \(\lambda = \dfrac{h}{\sqrt{2mE_K}}\); and for a charge \(q\) accelerated from rest through a potential difference \(V\), since \(E_K = qV\), \(\lambda = \dfrac{h}{\sqrt{2mqV}}\). This holds for any charged particle, but not when \(V\) is so large that the speed approaches that of light.
  4. For an electron accelerated through \(V\) volts, \[\lambda\ (\text{in nm}) = \frac{1.228}{\sqrt{V\ (\text{in volts})}} \qquad \text{--- (14.7)}\]
Davisson and Germer Experiment
  1. An electron beam from an electron gun was scattered by a nickel crystal in an evacuated chamber, and the scattered intensity was measured as a function of the scattering angle \(\theta\).
  2. The intensity was not uniform in all directions but showed peaks and troughs, that is, a diffraction pattern — a property only of waves.
  3. On varying the accelerating potential from 44 V to 68 V, a peak was observed at a scattering angle of 50º for 54 V. From Eq. (14.7), for \(V = 54\) V, \[\lambda = \frac{1.228}{\sqrt{54}} = 0.167\ \mathrm{nm} \qquad \text{--- (14.8)}\] while the diffraction measurements gave 0.165 nm. The close agreement verified the de Broglie relation.
  4. De Broglie received the Nobel Prize in 1929; Davisson, Germer and Thomson shared it in 1937.
Wave-Particle Duality of Matter
  1. All moving particles have an associated frequency and wave number, and all waves have an associated energy and momentum. Quantum behaviour becomes noticeable for microscopic objects of size \(\le 10^{-6}\) m, the scale being set by the small magnitude of \(h\).
  2. For macroscopic objects the de Broglie wavelength is far too small to matter: about \(6.63 \times 10^{-36}\) m for a 45 kg student running at 8 km/hr, and about \(3.32 \times 10^{-38}\) m for a 1200 kg car at 60 km/hr.
  3. For an electron of kinetic energy 100 eV, the wavelength is \(1.23 \times 10^{-10}\) m, or 1.228 Å — comparable to atomic dimensions, so wave effects are significant.
  4. Whenever wavelengths are small compared with the dimensions of slits or obstacles, or are not measurable, Newtonian mechanics may be used.
  5. In general, particle nature dominates during interaction with matter, while wave nature dominates during travel through space, particularly when the confinement is of the same order as the associated wavelength.
  6. Electrons can be freed from a metal by thermionic emission (heating to about 2000 ºC), field emission (fields of about \(10^{6}\) V/m at a metal tip), or photo-electron emission (ultraviolet or visible radiation).
  7. Because resolution improves as the probe wavelength decreases, electron beams of several keV, with \(\lambda_e \ll 1 \times 10^{-10}\) m, give electron microscopes a resolution several hundred times better than optical microscopes.

14.9 Objective Questions

  1. Choose the correct answer.
    1. A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
    1. zinc
    2. aluminum
    3. nickel
    4. potassium
    1. Polychromatic (containing many different frequencies) radiation is used in an experiment on photoelectric effect. The stopping potential
    1. will depend on the average wavelength
    2. will depend on the longest wavelength
    3. will depend on the shortest wavelength
    4. does not depend on the wavelength
    1. An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
    1. electron
    2. proton
    3. α-particle
    4. hydrogen atom
    1. If \(N_{Red}\) and \(N_{Blue}\) are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
    1. \(N_{Red} < N_{Blue}\)
    2. \(N_{Red} = N_{Blue}\)
    3. \(N_{Red} > N_{Blue}\)
    4. \(N_{Red} \approx N_{Blue}\)
    1. The equation \(E = pc\) is valid
    1. for all sub-atomic particles
    2. is valid for an electron but not for a photon
    3. is valid for a photon but not for an electron
    4. is valid for both an electron and a photon

14.10 Check Your Concepts

  1. Why does a threshold frequency exist at all? Explain, using the photon picture, why radiation below this frequency fails to eject electrons no matter how long it is allowed to fall on the metal.

  2. A certain metal shows photoelectric emission when blue light falls on it, but not when red light of much higher intensity is used. Predict what will happen if (a) ultraviolet radiation and (b) green light are used instead, and justify each prediction.

  3. The wave theory of light predicts a waiting period of several hours before photoemission begins, while experiments show emission within \(10^{-9}\) s. Explain the reasoning behind the wave-theory prediction, and state exactly which assumption of that reasoning is abandoned in Einstein’s picture.

  4. Why is the stopping potential in a photoelectric experiment independent of the intensity of the incident radiation, while the saturation current is not?

  5. Graphs of stopping potential against incident frequency are plotted for three different emitter metals and are found to be parallel straight lines with different intercepts. What physical quantity does the common slope represent, and what information is carried by the differing intercepts?

  6. The cathode of a photocell meant to work in visible light is made of an alkali metal rather than of tungsten or nickel. Explain the reason using the idea of work function.

  7. Why is the window of a photoelectric tube made of quartz rather than of ordinary glass, and why is the tube evacuated?

  8. Explain, in terms of photons and electrons, why the photocurrent first rises with the accelerating potential and then becomes constant. What decides the value at which it becomes constant?

  9. Photons are not deflected by electric or magnetic fields, but photoelectrons are. Explain why, and describe one experimental situation in which this difference is put to use.

  10. In Compton’s experiment, the scattered X-rays contain a wavelength longer than the incident one. Explain why the wave theory of radiation cannot account for this, and what the photon picture assumes instead.

  11. If Planck’s constant is allowed to approach zero in the Compton shift relation, the result predicted by wave theory is recovered. Why is such a limiting check considered a good test of a new theory?

  12. The rest mass of a photon is zero, yet it carries momentum. Explain how this is possible, and state whether the relation \(p = E/c\) may be applied to an electron.

  13. De Broglie’s hypothesis could not be tested using an ordinary slit cut in a metal sheet. Explain why a crystal had to be used instead, and what property of crystals makes them suitable.

  14. In an electron diffraction tube, the accelerating voltage between the cathode and the anode is increased. State whether the diffraction rings on the screen become larger or smaller, and justify your answer using the de Broglie relation.

  15. Why can an electron microscope resolve much finer detail than an optical microscope? State two further advantages of using an electron beam as a probe.

  16. A proton and an electron are accelerated from rest through the same potential difference. Without doing any calculation, state which of them will have the longer de Broglie wavelength, and explain your reasoning.

  17. Every moving object has a de Broglie wavelength, yet a cricket ball never shows diffraction. Explain, referring both to the magnitude of Planck’s constant and to the condition under which diffraction becomes observable.

  18. State the circumstances under which the particle nature of an entity dominates and those under which its wave nature dominates. Give one experiment from this chapter as evidence for each case.

  19. A student says that making a beam of red light brighter increases the energy of each of its photons. Identify the error in this statement and explain what actually increases.

  20. Explain why the photoelectrons emitted from a metal surface do not all have the same kinetic energy, even though every incident photon of a monochromatic beam carries exactly the same energy.

  21. Answer in brief.

  1. What is photoelectric effect?
  2. Can microwaves be used in the experiment on photoelectric effect?
  3. Is it always possible to see photoelectric effect with red light?
  4. Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.
  5. What do you understand by the term wave-particle duality? Where does it apply?
  1. Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.

  2. It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.

  3. Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.

  4. State the importance of Davisson and Germer experiment.

14.11 Practice with Numericals

(Take \(h = 6.63 \times 10^{-34}\) J s, \(c = 3 \times 10^{8}\) m s\(^{-1}\), \(e = 1.6 \times 10^{-19}\) C, mass of electron \(= 9.11 \times 10^{-31}\) kg unless stated otherwise.)

  1. Calculate the energy of a photon of wavelength 5000 Å, expressing your answer both in joule and in electron volt.

  2. A lamp radiates 60 W of light energy, all of wavelength 550 nm. How many photons does it emit per second?

  3. Using the work function of calcium given in Table 14.1, calculate its threshold frequency and threshold wavelength.

  4. Radiation of wavelength 3000 Å is incident on a sodium surface of work function 2.4 eV. Find the maximum kinetic energy of the emitted photoelectrons in eV and the stopping potential.

  5. In a photoelectric experiment, light of wavelength 2500 Å produces a stopping potential of 1.5 V. Calculate the work function of the emitter material in eV.

  6. The threshold frequency of a metal is \(5.5 \times 10^{14}\) Hz. Radiation of frequency \(7.5 \times 10^{14}\) Hz falls on its surface. Calculate the maximum kinetic energy of the photoelectrons and their maximum speed.

  7. A graph of stopping potential against incident frequency for a certain metal cuts the frequency axis at \(4.5 \times 10^{14}\) Hz. Determine the work function of the metal in eV and the stopping potential for incident radiation of frequency \(9.0 \times 10^{14}\) Hz.

  8. A photocell carries a saturation current of \(7.2\ \mu\)A. Calculate the number of photoelectrons reaching the anode per second.

  9. X-rays are scattered by free electrons and observed at a scattering angle of \(60^\circ\). Calculate the Compton shift, taking the mass of the electron as \(9.11 \times 10^{-31}\) kg.

  10. An X-ray photon has a wavelength of 1.0 Å. Calculate its energy in eV and its momentum.

  11. An electron is accelerated from rest through a potential difference of 200 V. Calculate its de Broglie wavelength in nanometre.

  12. An α-particle of mass \(6.64 \times 10^{-27}\) kg and charge \(+2e\) is accelerated from rest through a potential difference of 500 V. Calculate its de Broglie wavelength.

  13. Calculate the speed of an electron whose de Broglie wavelength is 0.5 Å, and find the kinetic energy of this electron in eV.

  14. A cricket ball of mass 0.15 kg moves with a speed of 30 m s\(^{-1}\). Calculate its de Broglie wavelength and comment on whether its wave nature can be detected.

  15. A thermal neutron has kinetic energy 0.025 eV. Taking the mass of the neutron as \(1.67 \times 10^{-27}\) kg, calculate its de Broglie wavelength and state whether it is suitable for the diffraction study of crystals whose atomic spacing is about 2 Å.

  16. A proton and an α-particle are accelerated from rest through the same potential difference. Taking the mass of the α-particle as four times that of the proton and its charge as twice that of the proton, find the ratio of their de Broglie wavelengths.

  17. A photon and an electron both have a wavelength of 1.5 Å. Calculate the energy of the photon in eV and the kinetic energy of the electron in eV, and comment on the large difference between the two values.

  18. Ultraviolet radiation of wavelength 2000 Å falls on a silver surface whose work function is 4.3 eV. Calculate the maximum kinetic energy of the photoelectrons in eV and the de Broglie wavelength associated with the fastest photoelectron.

  19. What will be the energy of each photon in monochromatic light of frequency \(5 \times 10^{14}\) Hz?

  20. Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately \(4.1 \times 10^{-15}\) V s. Given that the charge of an electron is \(1.6 \times 10^{-19}\) C, find the value of Planck’s constant \(h\).

  21. The threshold wavelength of tungsten is \(2.76 \times 10^{-5}\) cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than \(2.76 \times 10^{-5}\) cm. (b) What will be the maximum kinetic energy of electrons ejected in each of the following cases: (i) if ultraviolet radiation of wavelength \(\lambda = 1.80 \times 10^{-5}\) cm and (ii) radiation of frequency \(4 \times 10^{15}\) Hz is made incident on the tungsten surface.

  22. Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.

  23. Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?

  24. Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for the emitter?

Incident wavelength (in Å) 2536 3650
Stopping potential (in V) 1.95 0.5
  1. Calculate the wavelength associated with an electron, its momentum and speed (a) when it is accelerated through a potential of 54 V, (b) when it is moving with kinetic energy of 150 eV.

  2. The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of (i) their momenta (ii) their kinetic energies?

  3. Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle is an α-particle, what are the possibilities for the other particle?

  4. What is the speed of a proton having de Broglie wavelength of 0.08 Å?

  5. In nuclear reactors, neutrons travel with energies of \(5 \times 10^{-21}\) J. Find their speed and wavelength.

  6. Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum. State which of the two will have the longer wavelength in each case.