Chapter 11 · Dual Nature of Radiation and Matter

11.1 Electron Emission and the Photoelectric Effect

Electrons inside a metal are free to move but cannot leave it without a minimum energy, the work function. Light can supply that energy — and the way it does so, with a sharp threshold frequency and no delay, refuses to fit the wave picture of light.

A night-vision tube amplifies starlight a hundred-thousand-fold, a photomultiplier in a hospital PET scanner registers single flashes of light, and the solar panels on a satellite are slowly charged positive simply by sitting in sunshine. All three rely on one fact: light falling on a surface can knock electrons out of it. In 1887 Heinrich Hertz noticed that a spark jumped more easily across a gap when ultraviolet light shone on the metal electrodes; within a few years Wilhelm Hallwachs and Philipp Lenard had shown that the light was releasing negatively charged particles — electrons — from the metal.

This section asks two questions. First, what holds electrons inside a metal, and how much energy does it take to get one out? The answer is a property of each surface, the work function. Second, how does the emission of electrons by light — the photoelectric effect — depend on the brightness and the colour of the light? The experiments give clean, simple answers, and those answers turn out to be impossible to explain if light is only a wave. That puzzle is what Section 11.2 resolves.

Electron Emission and the Work Function

A metal contains a huge number of free electrons — about $10^{28}$ to $10^{29}$ per cubic metre — that wander through the lattice of positive ions (this is why metals conduct; see Section 3.1). Yet the electrons do not leak out of the surface. An electron that tries to leave finds an excess of positive ions behind it and is pulled back. To escape it must be given enough energy to overcome this attraction.

📘 Definition 11.1 — Work function

The work function $\phi_0$ of a surface is the minimum energy that must be supplied to an electron inside it to remove the electron from the surface. It depends on the material and on the state of its surface (cleanliness, oxide layers). Work functions are a few electron volts: $$1\ \text{eV}=1.6\times10^{-19}\ \text{J}.$$ An electron given energy $E$ greater than $\phi_0$ can leave with kinetic energy up to $E-\phi_0$; electrons deeper in the metal need more than $\phi_0$ and come out slower.

Recall from Section 2.1 that one electron volt is the energy an electron gains or loses in crossing a potential difference of $1$ V. Typical work functions are listed below. Notice that the alkali metals (caesium, potassium, sodium), whose single outer electron is loosely held, have the smallest values.

Metal	Cs	K	Na	Ca	Al	Cu	Ag	Pt
$\phi_0$ (eV)	$2.14$	$2.30$	$2.75$	$3.20$	$4.28$	$4.65$	$4.70$	$5.65$

Typical values for clean surfaces. Measured values differ by about $0.1$–$0.3$ eV between experiments because they are sensitive to the surface. We use this table throughout the chapter.

The energy needed to free an electron can be supplied in three ways.

📘 Definition 11.2 — The three kinds of electron emission

Thermionic emission: the metal is heated (to about $2000$ K or more for tungsten); some free electrons gain enough thermal energy to escape. Used in the filament of an X-ray tube and in old cathode-ray tubes.

Field emission: a very strong electric field (about $10^8$–$10^9$ V/m), usually at a sharp metal tip, pulls electrons out. Used in field-emission electron microscopes and the tip of a scanning tunnelling microscope.

Photoelectric emission: light (or other electromagnetic radiation) of suitable frequency falls on the surface and the electrons absorb energy from it.

✏️ Example 1 — Work functions in joules

(a) Express the work functions of potassium ($2.30$ eV) and platinum ($5.65$ eV) in joules. (b) What is the minimum energy needed to release from a potassium surface enough electrons to carry a total charge of $1.0\ \mu\text{C}$? (c) An electron just inside a potassium surface is given $3.00$ eV. What is the greatest kinetic energy it can have outside the metal?

💡 Show solution▶

(a) Convert Multiply by $1.6\times10^{-19}$ J/eV: potassium $2.30\times1.6\times10^{-19}=3.68\times10^{-19}$ J; platinum $5.65\times1.6\times10^{-19}=9.04\times10^{-19}$ J.

(b) Count Number of electrons $n=\dfrac{Q}{e}$ $=\dfrac{1.0\times10^{-6}}{1.6\times10^{-19}}$ $=6.25\times10^{12}$. Energy $=n\phi_0$ $=(6.25\times10^{12})(3.68\times10^{-19})$ $\approx2.3\times10^{-6}$ J.

(c) Leftover At most $3.00-2.30=0.70$ eV $=1.1\times10^{-19}$ J; less if the electron loses energy on its way to the surface.

Check In (b), $n\phi_0=\dfrac{Q}{e}\phi_0$ $=Q\times(2.30\ \text{V})$ $=1.0\times10^{-6}\times2.30$ J $=2.3\ \mu\text{J}$ ✓ — an energy in eV divided by $e$ is a potential in volts.

The Photoelectric Effect

Hallwachs's experiment is easy to picture. A clean zinc plate sits on top of a gold-leaf electroscope. If the plate is charged negatively and ultraviolet light shines on it, the leaves collapse: the plate is losing negative charge. If it is charged positively, nothing happens. An uncharged plate becomes slightly positive. Light is driving electrons out of the zinc, and a positive plate simply pulls them back.

📘 Definition 11.3 — Photoelectric effect and photoelectrons

The photoelectric effect is the emission of electrons from a surface (usually a metal) when electromagnetic radiation of high enough frequency falls on it. The electrons emitted are called photoelectrons, and the current they carry in a circuit is the photoelectric current (photocurrent).

The apparatus. To study the effect quantitatively we use an evacuated glass (or quartz, which transmits ultraviolet) tube containing two electrodes: a photosensitive plate $C$, the emitter, on which the light falls, and a plate $A$, the collector. A battery and a potential divider set the potential $V$ of the collector relative to the emitter — positive (accelerating) or negative (retarding) — and a sensitive microammeter measures the photocurrent. The light can be filtered to a single frequency $\nu$, and its intensity varied. The tube is evacuated so that the electrons do not collide with air molecules. Interactive 1 below is this apparatus.

Experimental Study of the Photoelectric Effect

Three things can be varied: the intensity of the light, the potential of the collector, and the frequency of the light (and, by changing the plate, the metal). Here is what is found.

1. Intensity (frequency and potential fixed). With the collector positive, the photocurrent is directly proportional to the intensity of the light. More intense light releases more photoelectrons per second.

2. Collector potential (frequency and intensity fixed). As $V$ is made more positive the current rises and then levels off: once every photoelectron emitted is being collected, raising $V$ further cannot increase the current. Now make $V$ zero and then negative. Even at $V=0$ there is a current — the electrons leave the emitter with kinetic energy and some reach the collector unaided. A negative collector repels them; only electrons with enough kinetic energy get through, and the current falls. At one particular negative potential, even the fastest photoelectrons are turned back and the current becomes zero.

📘 Definition 11.4 — Saturation current and stopping potential

The saturation current is the largest photocurrent for a given light: the current when every photoelectron emitted is collected. If $n$ electrons are emitted per second, $I_\text{sat}=ne$.

The stopping potential $V_0$ (cut-off potential) is the smallest negative potential of the collector, relative to the emitter, at which the photocurrent becomes zero. The fastest photoelectrons just fail to reach the collector, so by energy conservation (Section 2.1) $$K_{\max}=\tfrac12mv_{\max}^2=eV_0 .$$ A stopping potential of $V_0$ volts means $K_{\max}=V_0$ electron volts.

Repeat the potential sweep with brighter light of the same frequency: the saturation current is larger, but the current falls to zero at the same stopping potential. The maximum kinetic energy of the photoelectrons does not depend on the intensity.

3. Frequency (intensity adjusted so the saturation current is the same). Now the curves reach the same saturation level but cut off at different potentials: the higher the frequency, the larger the stopping potential. Plotting $V_0$ against $\nu$ gives a straight line. Extend it down to $V_0=0$: it meets the frequency axis at a frequency $\nu_0$. Below $\nu_0$ no photoelectrons are emitted at all, however intense the light and however long it shines. Different metals give different values of $\nu_0$ — but, remarkably, lines with the same slope.

4. Time. When light of frequency above $\nu_0$ is switched on, the emission begins at once — within about $10^{-9}$ s — even for extremely faint light.

📐 Law 11.1 — The experimental laws of photoelectric emission

For a given clean photosensitive surface:

(i) There is a threshold frequency $\nu_0$, characteristic of the surface, below which no photoelectric emission occurs, whatever the intensity of the light.

(ii) Above $\nu_0$, the photocurrent (number of photoelectrons emitted per second) is directly proportional to the intensity of the light.

(iii) The maximum kinetic energy $K_{\max}=eV_0$ of the photoelectrons is independent of the intensity, and increases linearly with the frequency of the light.

(iv) Emission is practically instantaneous (delay under $10^{-9}$ s), even for very faint light.

The threshold frequency corresponds to a threshold wavelength $\lambda_0=c/\nu_0$: emission requires $\nu\gt\nu_0$, that is $\lambda\lt\lambda_0$. For most metals $\lambda_0$ is in the ultraviolet; for the alkali metals it is in the visible.

✏️ Example 2 — From stopping potential to speed

In a photoelectric experiment the photocurrent falls to zero when the collector is at $-1.5$ V relative to the emitter. Find (a) the maximum kinetic energy of the photoelectrons in eV and in joules and (b) their maximum speed. ($m_e=9.1\times10^{-31}$ kg.)

💡 Show solution▶

(a) Energy $V_0=1.5$ V, so $K_{\max}=eV_0=1.5$ eV $=1.5\times1.6\times10^{-19}=2.4\times10^{-19}$ J.

(b) Speed $$v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}=\sqrt{\frac{2(2.4\times10^{-19})}{9.1\times10^{-31}}}\approx7.3\times10^{5}\ \text{m/s}.$$

Check $7.3\times10^5$ m/s is about $0.2\%$ of the speed of light, so the non-relativistic $\tfrac12mv^2$ is fine.

✏️ Example 3 — Counting photoelectrons

A small lamp, which radiates equally in all directions, shines on a photocell from $0.50$ m away. The saturation current is $4.0\ \mu\text{A}$ and the stopping potential is $0.60$ V. (a) How many photoelectrons are emitted per second? (b) The lamp is replaced by one of the same colour but twice the power. What are the new saturation current and stopping potential? (c) The original lamp is moved to $1.0$ m. Repeat (b).

💡 Show solution▶

(a) Rate $n=\dfrac{I_\text{sat}}{e}=\dfrac{4.0\times10^{-6}}{1.6\times10^{-19}}=2.5\times10^{13}$ electrons per second.

(b) Brighter Same frequency, twice the intensity: $I_\text{sat}=8.0\ \mu\text{A}$ (Law 11.1(ii)); $V_0=0.60$ V, unchanged (Law 11.1(iii)).

(c) Farther Intensity from a point source falls as $1/r^2$; doubling $r$ gives $\tfrac14$ the intensity: $I_\text{sat}=1.0\ \mu\text{A}$. Still $V_0=0.60$ V.

Exam tip "Intensity changes the current, frequency changes the stopping potential" — one line that wins many MCQs.

✏️ Example 4 — Partly stopped photoelectrons

Photoelectrons leave an emitter with kinetic energies from zero up to $3.0$ eV. What is the greatest kinetic energy with which an electron can arrive at the collector when the collector is at (a) $-1.8$ V, (b) $+2.0$ V, (c) $-3.5$ V relative to the emitter? In (b), what is the least kinetic energy on arrival?

💡 Show solution▶

Idea Crossing a potential difference $\Delta V$ changes an electron's kinetic energy by $-e\Delta V$: a collector at $+V$ adds $V$ eV, a collector at $-V$ removes $V$ eV.

(a) Retarding $3.0-1.8=1.2$ eV. Only electrons that left with more than $1.8$ eV get there, so the current is reduced but not zero.

(b) Accelerating Fastest: $3.0+2.0=5.0$ eV; slowest (left with nearly zero): $2.0$ eV.

✏️ Example 5 — Finding the threshold frequency from data

A student measures the stopping potential for one metal at four frequencies (illustrative data):

$\nu$ ($10^{14}$ Hz)	$6.0$	$7.0$	$8.0$	$9.0$
$V_0$ (V)	$0.35$	$0.77$	$1.18$	$1.60$

(a) Show that the points lie on a straight line and find its slope. (b) Find the threshold frequency and threshold wavelength. (c) Will light of wavelength $550$ nm release photoelectrons? If so, estimate the stopping potential.

💡 Show solution▶

(a) Slope Successive rises are $0.42,\ 0.41,\ 0.42$ V for each $1.0\times10^{14}$ Hz — constant, so a straight line. Using the end points, $$\text{slope}=\frac{1.60-0.35}{(9.0-6.0)\times10^{14}}\approx4.2\times10^{-15}\ \text{V s}.$$

(b) Threshold Extrapolate to $V_0=0$: $\nu_0=6.0\times10^{14}-\dfrac{0.35}{4.17\times10^{-15}}$ $\approx5.2\times10^{14}$ Hz. Then $\lambda_0=\dfrac{c}{\nu_0}=\dfrac{3.0\times10^8}{5.16\times10^{14}}\approx5.8\times10^{-7}$ m $=580$ nm.

(c) 550 nm $\nu=\dfrac{3.0\times10^8}{550\times10^{-9}}=5.45\times10^{14}$ Hz $\gt\nu_0$: yes. $V_0\approx(4.17\times10^{-15})$ $\times(5.45-5.16)\times10^{14}$ $\approx0.12$ V.

Look ahead The slope, about $4.1\times10^{-15}$ V s, is the same for every metal. Section 11.2 shows it equals $h/e$ — and so measures the Planck constant.

Choose a metal for the emitter, the wavelength and intensity of the light, and the collector potential $V$. The tube (left) shows photoelectrons leaving the emitter and being helped or held back by $V$; the graph (right) is photocurrent against $V$, with the dot at your setting. Press Keep curve, change one thing, and compare: intensity moves the saturation level, wavelength moves the stopping potential, and below the threshold frequency nothing happens at any intensity. The stopping potential and saturation current are computed (using Einstein's equation from Section 11.2, with $hc=1240$ eV nm; the saturation current is set to $10\ \mu\text{A}$ at full intensity); the rounded shape of the curve near $V=0$ is illustrative.

Metal:$\lambda$:350 nmintensity:60%$V$:+1.00 V

frequency $\nu$—

threshold $\nu_0$—

stopping potential $V_0$—

photocurrent at $V$—

The Photoelectric Effect and the Wave Theory of Light

By 1900 the wave theory of light was one of the great successes of physics: interference and diffraction (Chapter 10) and Maxwell's electromagnetic waves (Chapter 8) all fitted it perfectly. In the wave picture, light is a continuous electromagnetic wave whose energy is spread smoothly over the wavefront; its intensity is proportional to the square of the amplitude of its electric field. What would such a wave do to the electrons in a metal? The electric field would shake each electron, and the electron would steadily soak up energy until it had enough to escape. Compare each prediction with Law 11.1.

Prediction 1 — energy should grow with intensity. A more intense wave has a larger electric field and shakes the electrons harder, so the photoelectrons should come out faster: $K_{\max}$ should increase with intensity. Observed: $K_{\max}$ does not depend on intensity at all.

Prediction 2 — no threshold. Any frequency, even red or infrared light, should eventually release electrons if it is intense enough or shines long enough, because energy keeps accumulating. Observed: below $\nu_0$, nothing — not even with very intense light.

Prediction 3 — a time lag for faint light. The energy falling on the tiny area occupied by one electron is minute, so in faint light an electron would need minutes or hours to collect the work function. Observed: emission starts within $10^{-9}$ s. (Example 6 estimates the lag.)

There is also no reason in the wave picture why $K_{\max}$ should depend on frequency, let alone linearly. The wave theory explains the existence of photoemission — light carries energy — but not one of its quantitative features. Something about the way light delivers its energy must be different.

✏️ Example 6 — How long would a wave take?

Faint light of intensity $1.0\times10^{-3}\ \text{W/m}^2$ falls on a sodium surface ($\phi_0=2.75$ eV). Suppose, as the wave picture suggests, that each electron absorbs the energy falling on an area of about one atom, $1.0\times10^{-19}\ \text{m}^2$, and nothing else. Estimate how long an electron must wait before it can escape. Compare with experiment.

💡 Show solution▶

Power Power collected by one electron $P=IA$ $=(1.0\times10^{-3})(1.0\times10^{-19})$ $=1.0\times10^{-22}$ W.

Energy needed $\phi_0=2.75\times1.6\times10^{-19}=4.4\times10^{-19}$ J.

Time $t=\dfrac{\phi_0}{P}=\dfrac{4.4\times10^{-19}}{1.0\times10^{-22}}=4.4\times10^{3}$ s — about $73$ minutes.

Compare Experimentally, with light of frequency above $\nu_0$, the first photoelectrons appear within $10^{-9}$ s — more than a trillion times sooner. The energy is not spread thinly over the wave; it arrives in concentrated packets.

Two predictions side by side for the same metal and the same light. Left — wave theory: one electron soaks up the energy falling on an area of one atom ($1.0\times10^{-19}\ \text{m}^2$, an assumed size) until it has collected $\phi_0$; the clock runs fast (speed-up shown) so you can watch it. Any colour works if you wait. Right — experiment: light arrives in packets (dots). Above the threshold each packet can free an electron at once; below it, nothing ever happens. Change the intensity over twelve powers of ten and the colour, and see which picture changes.

Metal:Light:intensity:10⁻³ W/m²

wave-theory waiting time—

observed delay—

light frequency—

⚠️ Common mistakes with the photoelectric effect

1. "Brighter light gives faster electrons." No — intensity changes the number of photoelectrons (the current), never $K_{\max}$ or $V_0$.

2. Sign of the stopping potential. The collector is at $-V_0$ relative to the emitter; $V_0$ itself is quoted as a positive number, and $K_{\max}=eV_0$.

3. "All photoelectrons have energy $eV_0$." $eV_0$ is the maximum. Electrons from below the surface lose energy getting out, so energies range from $0$ to $K_{\max}$ — that is why the current falls gradually as the collector is made negative.

4. Threshold wavelength the wrong way round. Emission needs $\nu\gt\nu_0$, which means $\lambda\lt\lambda_0$: shorter wavelengths work, longer ones do not.

5. Saturation current and potential. Raising $V$ beyond saturation does not increase the current; only more light (more electrons per second) does.

Exercises

Work function and electron emission. Use $1\ \text{eV}=1.6\times10^{-19}$ J and the table of work functions.

1 Express the work function of caesium in joules.
2 The work function of a metal is $7.36\times10^{-19}$ J. Express it in eV.
3 Which metal in the table needs the least energy to release an electron? Suggest why the alkali metals have small work functions.
4 Name the kind of electron emission in each case: (a) a white-hot tungsten filament in an X-ray tube; (b) a needle tip in a field of $10^9$ V/m; (c) ultraviolet light on a zinc plate.
5 An electron just inside a sodium surface receives $3.50$ eV. What is the greatest kinetic energy it can have after leaving, in eV and in joules?
6 A zinc plate on an electroscope is charged (a) negatively, (b) positively, and then lit with ultraviolet light. Describe and explain what the leaves do in each case.
7 Find the minimum energy needed to release from potassium enough electrons to carry a total charge of $1.0$ mC.

💡 Show solutions▶

1 $2.14\times1.6\times10^{-19}\approx3.42\times10^{-19}$ J.

2 $\dfrac{7.36\times10^{-19}}{1.6\times10^{-19}}=4.60$ eV.

3 Caesium ($2.14$ eV). An alkali atom has a single outer electron far from the nucleus and well shielded by inner electrons, so it is weakly held.

4 (a) Thermionic, (b) field, (c) photoelectric emission.

5 $3.50-2.75=0.75$ eV $=0.75\times1.6\times10^{-19}=1.2\times10^{-19}$ J.

6 (a) The leaves collapse: photoelectrons leave the plate and are repelled by its negative charge, so the plate loses its charge. (b) No change: any photoelectron emitted is attracted straight back by the positive plate.

7 Energy $=\dfrac{Q}{e}\phi_0$ $=Q\times2.30\ \text{V}$ $=(1.0\times10^{-3})(2.30)$ $=2.3\times10^{-3}$ J.

Photocurrent, saturation and stopping potential. $m_e=9.1\times10^{-31}$ kg, $e=1.6\times10^{-19}$ C.

1 The stopping potential in an experiment is $0.80$ V. Find the maximum kinetic energy of the photoelectrons in eV and in joules.
2 Find the maximum speed of the photoelectrons in part 1.
3 The fastest photoelectrons from a surface move at $6.0\times10^5$ m/s. What is the stopping potential?
4 A saturation current of $2.4\ \mu\text{A}$ is measured. How many photoelectrons leave the emitter each second?
5 In part 4 the intensity of the light is tripled without changing its frequency. Give the new saturation current. What happens to the stopping potential?
6 The light in part 4 comes from a small lamp $0.50$ m away. The lamp is moved to $1.0$ m. Find the new saturation current.
7 Photoelectrons leave a surface with kinetic energies up to $2.0$ eV. With what greatest kinetic energy do they reach a collector at (a) $-1.2$ V and (b) $+1.2$ V relative to the emitter?
8 Explain why the photocurrent is not zero when the collector and emitter are at the same potential.

💡 Show solutions▶

1 $K_{\max}=eV_0=0.80$ eV $=1.28\times10^{-19}\approx1.3\times10^{-19}$ J.

2 $v=\sqrt{\dfrac{2(1.28\times10^{-19})}{9.1\times10^{-31}}}\approx5.3\times10^5$ m/s.

3 $K=\tfrac12(9.1\times10^{-31})(6.0\times10^5)^2$ $\approx1.64\times10^{-19}$ J; $V_0=\dfrac{K}{e}\approx1.0$ V.

4 $n=\dfrac{2.4\times10^{-6}}{1.6\times10^{-19}}=1.5\times10^{13}$ per second.

5 $I_\text{sat}\propto$ intensity: $7.2\ \mu\text{A}$. The stopping potential is unchanged.

6 Intensity $\propto1/r^2$ falls to $\tfrac14$: $0.60\ \mu\text{A}$.

7 (a) $2.0-1.2=0.80$ eV; (b) $2.0+1.2=3.2$ eV.

8 Photoelectrons leave the emitter with kinetic energy, so some cross the gap and reach the collector even without an accelerating potential.

III

Frequency, threshold and graphs. $c=3.0\times10^8$ m/s.

1 Sketch, on one set of axes, photocurrent against collector potential for light of one frequency at two intensities. On a second set, sketch it for two frequencies $\nu_1\lt\nu_2$ with intensities adjusted to give the same saturation current. Label $V_0$ on each curve.
2 The graph of stopping potential against frequency for a metal passes through $(7.0\times10^{14}\ \text{Hz},\ 0.90\ \text{V})$ and $(9.0\times10^{14}\ \text{Hz},\ 1.73\ \text{V})$. Find its slope and the threshold frequency.
3 Find the threshold wavelength of the metal in part 2.
4 A metal has threshold frequency $5.5\times10^{14}$ Hz. Will light of wavelength (a) $500$ nm and (b) $600$ nm release photoelectrons? Does the answer to (b) change if the light is made very intense?
5 The threshold wavelength of a surface is $540$ nm. Which of $480$ nm and $650$ nm light can cause photoemission?
6 Predict the stopping potential for the metal of part 2 when light of frequency $1.1\times10^{15}$ Hz is used.
7 Explain why the photocurrent–potential curves for different intensities of the same light all reach zero at the same potential.

💡 Show solutions▶

1 First graph: two curves with different saturation levels (the brighter higher), meeting the $V$-axis at the same point $-V_0$. Second graph: same saturation level; the $\nu_2$ curve cuts off at a more negative potential, $V_{02}\gt V_{01}$.

2 Slope $=\dfrac{1.73-0.90}{2.0\times10^{14}}\approx4.15\times10^{-15}$ V s. $\nu_0=7.0\times10^{14}-\dfrac{0.90}{4.15\times10^{-15}}$ $\approx4.8\times10^{14}$ Hz.

3 $\lambda_0=\dfrac{3.0\times10^8}{4.83\times10^{14}}\approx6.2\times10^{-7}$ m $=620$ nm.

4 (a) $\nu=6.0\times10^{14}$ Hz $\gt\nu_0$: yes. (b) $\nu=5.0\times10^{14}$ Hz $\lt\nu_0$: no — and no amount of intensity changes that.

5 Only $480$ nm (shorter than $\lambda_0$, so higher frequency than $\nu_0$).

6 $V_0=0.90+(4.15\times10^{-15})$ $\times(1.1-0.70)\times10^{15}$ $\approx0.90+1.66\approx2.6$ V.

7 The current becomes zero when even the fastest electrons are stopped, i.e. at $V_0=K_{\max}/e$. $K_{\max}$ depends on the frequency and the metal but not on the intensity, so all the curves cut off at the same $-V_0$.

The wave picture and applications.

1 State three predictions of the wave theory of light about the photoelectric effect, and the observation that contradicts each.
2 Light of intensity $5.0\times10^{-3}\ \text{W/m}^2$ falls on potassium. Assuming that each electron absorbs the energy falling on $1.0\times10^{-19}\ \text{m}^2$, estimate the wave-theory waiting time before emission.
3 Repeat part 2 for starlight-level illumination, $1.0\times10^{-9}\ \text{W/m}^2$. Give the answer in years.
4 Modelling. A "solar-blind" ultraviolet flame detector uses a surface with threshold wavelength $280$ nm, so that it ignores sunlight filtered by the atmosphere. Find its threshold frequency, and show that no visible light ($400$–$700$ nm) can trigger it, however bright.
5 A light meter's photocurrent is $12$ nA. How many photoelectrons are emitted in one minute?
6 Why is the observation that emission begins within $10^{-9}$ s, even in faint light, such strong evidence against the wave picture?

💡 Show solutions▶

1 (i) $K_{\max}$ should rise with intensity — it does not. (ii) Any frequency should work given enough intensity or time — there is a sharp threshold $\nu_0$. (iii) Faint light should show a time lag — emission is instantaneous.

2 $P=(5.0\times10^{-3})(1.0\times10^{-19})=5.0\times10^{-22}$ W; $t=\dfrac{3.68\times10^{-19}}{5.0\times10^{-22}}\approx7.4\times10^2$ s $\approx12$ min.

3 $P=1.0\times10^{-28}$ W; $t=3.68\times10^{9}$ s $\approx\dfrac{3.68\times10^9}{3.16\times10^7}\approx1.2\times10^2$ years.

4 $\nu_0=\dfrac{3.0\times10^8}{280\times10^{-9}}\approx1.07\times10^{15}$ Hz. The highest visible frequency (at $400$ nm) is $7.5\times10^{14}$ Hz $\lt\nu_0$, so visible light never causes emission.

5 $n=\dfrac{It}{e}=\dfrac{(12\times10^{-9})(60)}{1.6\times10^{-19}}=4.5\times10^{12}$.

6 In the wave picture the energy reaching one electron in faint light is so small that it would take minutes to years (parts 2–3) to collect $\phi_0$. Instant emission means the energy is delivered in concentrated amounts, not spread over the wavefront.

🎯 Exam corner — electron emission and the photoelectric effect BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Define (a) stopping potential and (b) threshold frequency. (c) Draw graphs of photocurrent against collector potential for light of the same frequency but two different intensities, and state what the graphs show.

💡 Show solution▶

(a) The minimum negative potential of the collector relative to the emitter at which the photocurrent becomes zero; $eV_0=K_{\max}$.

(b) The minimum frequency of incident radiation below which no photoelectrons are emitted from a given surface, whatever the intensity.

(c) Two curves rising to different saturation currents (higher for the more intense light) and meeting the potential axis at the same point $-V_0$. They show that the photocurrent is proportional to intensity, but the stopping potential (hence $K_{\max}$) is independent of intensity.

Exam tip Label the axes, mark $-V_0$ on the negative side, and mark both saturation levels — the marks are for the labels.

E2. Board · 1 mark Assertion (A): For a given metal, the stopping potential increases when the incident light is made brighter. Reason (R): Brighter light has a larger electric-field amplitude. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

💡 Show solution▶

Concept The stopping potential does not depend on intensity, so A is false. Intensity is proportional to the square of the field amplitude, so R is true — it is exactly the wave-theory argument that fails.

Answer (D). Exam tip: a true statement of wave theory used to "explain" a false claim is a favourite trap.

E3. JEE Main The graph of stopping potential against frequency for a metal passes through $(5.0\times10^{14}\ \text{Hz},\ 0.30\ \text{V})$ and $(8.0\times10^{14}\ \text{Hz},\ 1.54\ \text{V})$. The threshold frequency of the metal is about (A) $3.6\times10^{14}$ Hz (B) $4.3\times10^{14}$ Hz (C) $5.0\times10^{14}$ Hz (D) $5.7\times10^{14}$ Hz.

💡 Show solution▶

Slope $\dfrac{1.54-0.30}{3.0\times10^{14}}\approx4.13\times10^{-15}$ V s.

Extrapolate $\nu_0=5.0\times10^{14}-\dfrac{0.30}{4.13\times10^{-15}}$ $\approx5.0\times10^{14}-0.73\times10^{14}$ $\approx4.3\times10^{14}$ Hz.

Answer (B). Exam tip: (D) comes from adding $0.73\times10^{14}$ Hz instead of subtracting it; check that $\nu_0$ is below the lowest measured frequency.

E4. NEET In a photoelectric experiment the intensity of the incident light is doubled while its frequency is kept the same. Then (A) the saturation current doubles and the stopping potential is unchanged (B) both the saturation current and the stopping potential double (C) the saturation current is unchanged and the stopping potential doubles (D) the saturation current doubles and the stopping potential halves.

💡 Show solution▶

Concept Twice as many photoelectrons per second, each with the same range of energies.

Answer (A). Exam tip: only frequency (and the metal) can change $V_0$.

E5. JEE Advanced (One or more correct.) Light of frequency $\nu$ greater than the threshold frequency $\nu_0$ falls on a metal surface in a photoelectric tube. Which statements are correct? (A) The photocurrent is not zero when the collector is at the same potential as the emitter. (B) The saturation current is proportional to the intensity of the light. (C) For the same light, the stopping potential would be different for a different metal. (D) If $\nu$ is reduced below $\nu_0$ and the intensity is increased tenfold, photoemission continues.

💡 Show solution▶

Options (A) True: photoelectrons have kinetic energy and some reach the collector unaided. (B) True: Law 11.1(ii). (C) True: different metals have different threshold frequencies, so different $V_0$ at the same $\nu$ (parallel $V_0$–$\nu$ lines). (D) False: below $\nu_0$ there is no emission at any intensity.

Answer (A), (B), (C). Exam tip: in multi-correct questions test each statement against one law at a time.

E6. Board · Case study A school physics club studies photoemission from a metal plate in an evacuated tube. With the collector strongly positive the current is $3.2\ \mu\text{A}$; as the collector is made negative the current falls and becomes zero at $-0.90$ V. (i) (2 marks) Find the number of photoelectrons emitted per second and their maximum kinetic energy in joules. (ii) (1 mark) The intensity of the light is halved. What are the new saturation current and stopping potential? (iii) (1 mark) Find the maximum speed of the photoelectrons.

💡 Show solution▶

(i) $n=\dfrac{3.2\times10^{-6}}{1.6\times10^{-19}}=2.0\times10^{13}$ per second; $K_{\max}=eV_0=0.90$ eV $=1.44\times10^{-19}\approx1.4\times10^{-19}$ J.

(ii) $1.6\ \mu\text{A}$; the stopping potential stays $0.90$ V.

(iii) $v_{\max}=\sqrt{\dfrac{2(1.44\times10^{-19})}{9.1\times10^{-31}}}\approx5.6\times10^5$ m/s. Exam tip: keep $K$ in joules (not eV) inside $\sqrt{2K/m}$.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why do satellites in sunlight charge themselves up, and which surface materials make the effect worse?

Why it matters. Ultraviolet sunlight releases photoelectrons from every sunlit surface of a spacecraft. The craft loses electrons and becomes positively charged until it pulls its slowest photoelectrons back. Differences in charging between sunlit and shaded parts can cause electrostatic discharges that damage electronics — a design problem for every space agency, from ESA and NASA to ISRO and JAXA, and a reason why scientific probes that measure space plasma must control their own potential.

What to do.

For aluminium, copper, silver and platinum (table above), find the threshold wavelength using $\lambda_0=\dfrac{hc}{\phi_0}$ with $hc=1240$ eV nm (Section 11.2 justifies this; take it on trust here). Which part of the spectrum is each in?
Sunlight in space contains a strong ultraviolet line at $121.6$ nm, whose packets of energy carry $1240/121.6$ eV each. Find the maximum kinetic energy of photoelectrons it releases from aluminium.
A spacecraft charges positive until its potential reaches roughly the stopping potential of these fastest photoelectrons. Estimate that potential. Why does charging stop there?
Ordinary visible sunlight is far more intense than this ultraviolet line. Explain, using Law 11.1, why the visible light contributes almost nothing to the charging.

Tools: Calculator; the work-function table; optionally a published solar spectrum to see how much energy lies below $290$ nm.

Hand in: A table of threshold wavelengths, the charging-potential estimate and a short paragraph recommending a coating for a sensor that must not charge up.

Going further. Engineers coat spacecraft with slightly conducting materials so that charge spreads evenly. Explain why making the whole surface one equipotential (Section 2.2) removes the danger even though the craft is still charged.

Instructor note: $\lambda_0$: Al $1240/4.28\approx290$ nm, Cu $\approx267$ nm, Ag $\approx264$ nm, Pt $\approx219$ nm — all ultraviolet. $121.6$ nm gives $10.2$ eV, so $K_{\max}\approx10.2-4.28\approx5.9$ eV on aluminium and the craft charges to a few volts positive (about $+6$ V at most from this line); beyond that the photoelectrons are pulled back and the net current vanishes. Visible light is below threshold for all these metals, so its intensity is irrelevant (Law 11.1(i)). Real charging in sunlight is typically a few volts positive; dangerous differential charging arises between sunlit and shaded insulators.

🧪 Self-check quiz

0/5 correct

Q1. For a given photosensitive surface, the stopping potential depends on

Q2. The work function of sodium, $2.75$ eV, in joules is

Q3. A saturation current of $1.6\ \mu\text{A}$ means that the number of photoelectrons emitted per second is

Q4. Which observation can the wave theory of light not explain?

Q5. The photocurrent has reached its saturation value. If the collector potential is made still more positive, the current

← Chapter 11 — Dual Nature of Radiation and Matter 11.2 Einstein's Photoelectric Equation and the Photon →