Introduction and Photoelectric Effect

Q: What is the photoelectric effect?

It is the emission of electrons from a metal surface when light of frequency above the threshold frequency falls on it. The emitted electrons are called photoelectrons. Einstein explained it by treating light as photons of energy hν, each absorbed by a single electron.

Q: What are work function and threshold frequency?

Work function is the minimum energy an electron needs to escape from a metal surface, for example 2.14 eV for caesium. Threshold frequency is the smallest light frequency that can cause emission, ν₀ = φ₀/h. Below it no electrons come out, however bright the light.

Q: Why does maximum kinetic energy not depend on intensity?

Each electron absorbs a single photon, whose energy hν depends only on frequency. Brighter light means more photons, so more electrons are emitted, but each still gets the same hν. So Kmax = hν - φ₀ and the stopping potential stay unchanged.

Q: What is stopping potential?

It is the minimum negative potential of the collector, relative to the emitter, that just stops the fastest photoelectrons, making the photocurrent zero. It measures their maximum kinetic energy: eV₀ = Kmax. It depends on frequency and metal, not on intensity.

Q: Why does the photocurrent become saturated?

As the collector is made more positive it attracts more of the emitted electrons. Once every emitted electron reaches the collector, raising the potential further cannot increase the current. This saturation current depends only on how many electrons are emitted per second, which is set by the intensity.

Q: Why do photoelectrons have a range of kinetic energies?

Free electrons in a metal have different energies, and electrons deeper inside lose energy before reaching the surface. Only the least tightly bound electron escapes with hν - φ₀; the others come out slower. So energies range from zero up to Kmax.

Q: Which photoelectric effect questions come in NEET?

NEET often asks graph-based questions (photocurrent against intensity or potential, stopping potential against frequency), the effect of doubling intensity or frequency, threshold wavelength, and simple numericals on maximum kinetic energy and stopping potential. Working in electronvolts with 1240 divided by the wavelength in nanometres saves time.

Q: How is the photoelectric effect tested in JEE Main and Advanced?

JEE Main asks two-wavelength problems that remove the work function, stopping potential from graphs, photon counting and saturation current. JEE Advanced adds photoelectrons in magnetic fields, radiation pressure at oblique incidence, charging of an isolated emitter and combined problems with Bohr's model.

PhysicsDual Nature Of Radiation And MatterFor NEET aspirants

The photoelectric effect is the emission of electrons from a metal when light above a certain threshold frequency falls on it. It proved that light gives its energy in packets (photons) of $h ν$ : one photon frees one electron, so $K_{ma x} = h ν - ϕ_{0}$ . Intensity decides how many electrons come out; frequency decides how fast they come out. This page covers work function, the experiment and its graphs, Einstein's equation, photons and every standard numerical type. The photoelectric effect is one of the highest-scoring Modern Physics topics in JEE Main and NEET.

On this page1Work function2The experiment3Results and graphs4Why waves fail5Einstein's equation6Photons7Problem types8Revision map

Key Formulas - Quick Reference

★ Must learn Photon energy: $E = h ν = \frac{h c}{λ}$ ; shortcut $E (eV) = \frac{1240}{λ ( nm )}$
★ Must learn Einstein's equation: $K_{ma x} = h ν - ϕ_{0} = h (ν - ν_{0})$
★ Must learn Stopping potential: $e V_{0} = K_{ma x} = \frac{1}{2} m v_{ma x}^{2}$ , so $V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}$ (slope $\frac{h}{e}$ for every metal)
Threshold: $ν_{0} = \frac{ϕ _{0}}{h}$ , $λ_{0}$ $= \frac{h c}{ϕ _{0}}$ $= \frac{1240}{ϕ _{0} ( eV )} nm$ ; emission only if $ν \geq ν_{0}$ ( $λ \leq λ_{0}$ )
★ Must learn Photons per second: $N = \frac{P}{h ν} = \frac{P λ}{h c}$ ; photon momentum $p = \frac{h}{λ} = \frac{E}{c}$
Saturation current: $i_{s a t} = η N e$ ( $η$ = electrons emitted per incident photon)
Force of light: $F = \frac{P}{c}$ (absorbed), $F = \frac{2 P}{c}$ (perfectly reflected)
★ Must learn Intensity changes only the number of electrons ( $i_{s a t}$ ); frequency changes only their energy ( $K_{ma x}$ , $V_{0}$ )
Useful values: $h$ $= 6.63 \times 1 0^{- 34} J s$ $= 4.14 \times 1 0^{- 15} eV s$ , $1 eV = 1.6 \times 1 0^{- 19} J$ , $\frac{h}{e} = 4.14 \times 1 0^{- 15} V s$

1. Electron Emission and Work Function

A metal has free electrons, but they cannot leave on their own: if one tries, the positive ions pull it back. It must be given a minimum energy, the work function $ϕ_{0}$ , to escape.

Figure 1: The work function

ϕ_{0}

is the least energy that lets the most loosely held free electron leave the metal. It ranges from

2.14 eV

(Cs, lowest) to

5.65 eV

(Pt, highest) and depends on the metal and its surface.

Work function

ϕ_{0}

: the minimum energy needed by an electron to escape from a metal surface. It is measured in electronvolts: $1 eV = 1.6 \times 1 0^{- 19} J$ , the energy an electron gains across

1 V

. It depends on the metal and on its surface condition, never on the light used.

Type of emission	Energy supplied by	Example
Thermionic	heating the metal	filament of an electron gun (X-ray tube, old TV tube)
Field emission	a very strong electric field, about $1 0^{8} V m^{- 1}$	spark plug, sharp field-emission tips
Photoelectric	light of frequency $ν \geq ν_{0}$	photocell, solar cell, light meter

Metal	Cs	K	Na	Ca	Mo	Cu	Ag	Ni	Pt
$ϕ_{0}$ (eV)	2.14	2.30	2.75	3.20	4.17	4.65	4.70	5.15	5.65

Lowest: caesium ( $2.14 eV$ ); highest in the table: platinum ( $5.65 eV$ ).
Alkali metals (Li, Na, K, Rb, Cs) have low $ϕ_{0}$ and emit even with visible light. Zn, Cd, Mg and Cu need ultraviolet.

2. The Photoelectric Experiment

Light falls on the emitter plate C inside an evacuated tube; emitted electrons (photoelectrons) move to the collector A and a current flows in the circuit. The tube is evacuated so that electrons do not collide with gas molecules.

Figure 2: The set-up. Three knobs are varied one at a time: the intensity of light (distance of S), its frequency (filters) and the collector potential

V

(divider, commutator). The microammeter reads the photocurrent

i

Photocurrent $i$ : current due to photoelectrons reaching A; it is proportional to the number reaching A per second.
Saturation current $i_{s a t}$ : the maximum photocurrent, reached when A is positive enough to collect every emitted electron.
Stopping (cut-off) potential $V_{0}$ : the minimum negative potential of A (relative to C) at which the photocurrent becomes zero. It stops even the fastest electron, so $e V_{0} = K_{ma x}$ .

Scientist (year)	One-line fact for MCQs
Hertz (1887)	discovered the effect: UV light made sparks jump more easily across his detector
Hallwachs (1888)	a negatively charged zinc plate lost its charge in UV; a neutral plate became positive
Lenard	UV on the emitter of an evacuated tube gave a current; no emission below a threshold frequency
Einstein (1905)	explained it with light quanta (photons); Nobel Prize 1921
Millikan (1916)	verified Einstein's equation and measured $h$ from the $V_{0}$ - $ν$ slope; Nobel Prize 1923

3. The Results and Their Graphs

3.1 Effect of intensity

Keep the frequency fixed and vary the intensity (move the source closer or farther). The saturation current grows in proportion to intensity, but the curves all end at the same stopping potential.

Figure 3: Same frequency, three intensities. Saturation current

\propto

intensity (more photons, more electrons); the stopping potential

V_{0}

, and so

K_{ma x} = e V_{0}

, does not change.

3.2 Effect of collector potential

Start with A positive: the current rises and then saturates. Make A negative (retarding): only faster electrons can reach A against the field, so the current falls, and it becomes zero at $- V_{0}$ . The current is not zero at $V = 0$ , because electrons leave C with some kinetic energy and many still reach A.

Saturation current

Depends on intensity (number of photons per second), not on frequency (for the same photon flux). Set by how many electrons are emitted.

Stopping potential

Depends on frequency and the metal, not on intensity. Set by the energy of the fastest electron: $e V_{0} = K_{ma x}$ .

3.3 Effect of frequency

Keep the intensity fixed and change the frequency. The higher the frequency, the larger the stopping potential; the saturation current stays the same.

Figure 4: Same saturation current, three frequencies. Higher

ν

gives faster electrons, so a larger stopping potential is needed:

V_{03} > V_{02} > V_{01}

A plot of $V_{0}$ against $ν$ for any metal is a straight line. Below a threshold frequency $ν_{0}$ there is no emission at all, however intense the light or however long it shines. Different metals give parallel lines.

Figure 5:

V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}

. Parallel lines of slope

\frac{h}{e} = 4.14 \times 1 0^{- 15} V s

for every metal; the

ν

-intercept is the threshold

ν_{0}

(

5.2

6.6

11.2 \times 1 0^{14} Hz

for Cs, Na, Cu) and the extended line meets the

V_{0}

axis at

- \frac{ϕ _{0}}{e}

★ Must learn Laws of photoelectric emission

For a given metal there is a threshold frequency $ν_{0}$ ; below it no electrons are emitted, whatever the intensity.
Above $ν_{0}$ , photocurrent (saturation current) is proportional to intensity.
$K_{ma x}$ (and $V_{0}$ ) increases linearly with frequency and does not depend on intensity.
Emission is instantaneous: the time lag is less than about $1 0^{- 9} s$ , even for very dim light.

Figure 6: The six graphs to recognise at a glance. Intensity controls only the number of electrons (a); energy depends only on frequency (b, c, d).

K_{ma x}

against

λ

is a curve, not a line (e). Emission starts within about

1 0^{- 9} s

(f).

Exam Trick

Read the axes first. Anything plotted against intensity is either a line through O ( $i_{s a t}$ , number of electrons) or a horizontal line ( $K_{ma x}$ , $V_{0}$ , $ν_{0}$ ). Anything energy-like plotted against $ν$ or $\frac{1}{λ}$ is a straight line that starts at the threshold; against $λ$ it is a curve.

Key idea

Intensity decides how many electrons (saturation current); frequency decides how fast (stopping potential).

Quick Recall: tap to check

Intensity is doubled at the same frequency. What happens to

i_{s a t}

and

V_{0}

i_{s a t}

doubles;

V_{0}

is unchanged.

Why is the photocurrent not zero when

V = 0

Electrons leave the emitter with kinetic energy, so many reach the collector even without an accelerating field.

What is the slope of the

V_{0}

ν

graph, and does it depend on the metal?

\frac{h}{e} = 4.14 \times 1 0^{- 15} V s

, the same for every metal.

4. Why the Wave Theory Fails

In the wave picture, light energy arrives continuously and is shared by all the electrons on the surface. Every prediction of that picture contradicts experiment.

Figure 7: Wave picture: energy arrives continuously and is shared, so a dim beam would need hours to free an electron. Photon picture: energy arrives in lumps of

h ν

, each taken by a single electron at once.

Observation	Wave theory predicts	Photon theory explains
$K_{ma x}$ independent of intensity	brighter light, bigger field amplitude, so larger $K_{ma x}$	$K_{ma x}$ is set by one photon's energy $h ν$
Threshold $ν_{0}$ exists	any frequency works if the light is intense enough	a photon with $h ν < ϕ_{0}$ cannot free an electron
$K_{ma x}$ rises with $ν$	no reason for frequency to matter	$K_{ma x} = h ν - ϕ_{0}$
No time lag	dim light needs a long time to build up energy	one photon is absorbed at once

How long would the wave picture need? For light of $1 0^{- 5} W m^{- 2}$ on $2 cm^{2}$ of sodium ( $ϕ_{0} \approx 2 eV$ ), sharing the energy among the atoms of the top five atomic layers ( $\approx 8.6 \times 1 0^{15}$ electrons) gives about $1.4 \times 1 0^{6} s$ , roughly two weeks per electron. Experiment shows emission within $1 0^{- 9} s$ .

5. Einstein's Photoelectric Equation

Einstein (1905) proposed that light energy comes in quanta (photons) of energy $h ν$ , and that one photon is absorbed by one electron. Energy conservation for that single event gives the equation:

The electron receives the photon's whole energy $h ν$ (a photon cannot be absorbed in part).
It spends at least $ϕ_{0}$ to escape; the least tightly bound electron spends exactly $ϕ_{0}$ .
The rest is kinetic energy, largest for that least-bound electron.

K_{ma x} = \frac{1}{2} m v_{ma x}^{2} = h ν - ϕ_{0} = h (ν - ν_{0}) = e V_{0}

This is Einstein's photoelectric equation. Emission needs $K_{ma x} \geq 0$ , so $h ν \geq ϕ_{0}$ : $ν_{0} = \frac{ϕ _{0}}{h}$ , $λ_{0} = \frac{h c}{ϕ _{0}}$ .

Figure 8: One photon, one electron. The least tightly held electron (A) leaves with

K_{ma x} = h ν - ϕ_{0} = 3.10 - 2.14 = 0.96 eV

; deeper electrons (B) leave slower, which is why photoelectrons have a range of energies from

0

K_{ma x}

Dividing by $e$ gives a straight line in $ν$ :

V_{0} = \frac{h}{e} ν - \frac{ϕ _{0}}{e}

Slope $= \frac{h}{e}$ (same for every metal), $ν$ -intercept $= ν_{0}$ , $V_{0}$ -intercept $= - \frac{ϕ _{0}}{e}$ . Millikan used exactly this graph to measure $h$ and confirm the equation (Figure 9 shows the method with rubidium data).

Figure 9: Real-style data (rubidium, mercury lines). Least-squares slope

= 4.108 \times 1 0^{- 15} V s

gives

h

= e \times slope

= 6.57 \times 1 0^{- 34} J s

; threshold

ν_{0} = 5.10 \times 1 0^{14} Hz

;

ϕ_{0} = 2.09 eV

5.1 Threshold wavelength

Since $E (eV) = \frac{1240}{λ ( nm )}$ , a metal emits only for $λ \leq λ_{0}$ $= \frac{1240}{ϕ _{0} ( eV )} nm$ . Visible photons carry $1.8$ to $3.3 eV$ , so only low work-function metals respond to visible light.

Figure 10: A metal emits only if the photon energy line is above its work function, that is for

λ < λ_{0}

= \frac{1240}{ϕ _{0} ( eV )} nm

. Visible light (

1.8

3.3 eV

) works for Cs (

579 nm

) and Na (

451 nm

) but never for Cu or Pt, which need ultraviolet.

Exam Trick

Work in electronvolts. $K_{ma x} (eV)$ $= \frac{1240}{λ ( nm )} - ϕ_{0} (eV)$ , and the stopping potential in volts is the same number: $K_{ma x} = 1.2 eV$ means $V_{0} = 1.2 V$ . Convert to joules only when you need a speed: $v_{ma x} = \frac{2 K _{ma x}}{m}$ .

Key idea

One photon, one electron:

K_{ma x} = h ν - ϕ_{0}

. The

V_{0}

ν

line has slope

\frac{h}{e}

and starts at

ν_{0}

6. Photons: Energy, Momentum and Number

The photoelectric effect shows that light, when it exchanges energy with matter, behaves as a stream of particles called photons.

Energy $E = h ν = \frac{h c}{λ}$ ; momentum $p = \frac{h ν}{c} = \frac{h}{λ}$ ; speed $c$ in vacuum; rest mass zero.
All photons of one frequency have the same energy and momentum, whatever the intensity. A brighter beam has more photons, not stronger ones.
Photons are electrically neutral: electric and magnetic fields do not deflect them.
In a photon-particle collision, total energy and momentum are conserved, but the number of photons need not be: photons can be absorbed or created.

Counting photons. A source of power $P$ emits $N = \frac{P}{h ν} = \frac{P λ}{h c}$ photons per second. Intensity (power per unit area) $I = \frac{N h ν}{A}$ , so the photon flux is $\frac{I}{h ν}$ . If a fraction $η$ of the photons eject electrons, $i_{s a t} = η N e$ .

Light exerts a force. Each photon carries momentum, so a beam pushes on whatever absorbs or reflects it.

Figure 11: Light pushes. Photons per second

N = \frac{P}{h ν}

, each carrying

p = \frac{h ν}{c}

, so the force is

N p = \frac{P}{c}

when absorbed and

\frac{2 P}{c}

when reflected. A

1 W

beam fully absorbed pushes with only

3.3 \times 1 0^{- 9} N

JEE Advanced

Oblique incidence. A beam of intensity $I$ strikes area $A$ at angle $θ$ to the normal. Power intercepted $= I A cos θ$ , momentum per second along the beam $= \frac{I A cos θ}{c}$ .

Perfect absorber: normal force $\frac{I A cos ^{2} θ}{c}$ , plus a tangential force $\frac{I A sin θ cos θ}{c}$ .
Perfect reflector: normal force $\frac{2 I A cos ^{2} θ}{c}$ , no tangential force.
Reflectivity $r$ : normal force $\frac{( 1 + r ) I A cos ^{2} θ}{c}$ , tangential $\frac{( 1 - r ) I A sin θ cos θ}{c}$ ; radiation pressure $\frac{( 1 + r ) I cos ^{2} θ}{c}$ .

Isolated emitter. A metal sphere (radius $R$ ) that loses photoelectrons becomes positive. Emission stops when its potential reaches $V_{0}$ , so it gains at most $Q = 4 π ε_{0} R V_{0}$ .

Quick Recall: tap to check

How many photons per second does a

1 mW

620 nm

laser emit?

E

= \frac{1240}{620}

= 2 eV

= 3.2 \times 1 0^{- 19} J

, so

N

= \frac{1 0 ^{- 3}}{3.2 \times 1 0 ^{- 19}}

= 3.1 \times 1 0^{15} s^{- 1}

Does a photon have mass or charge?

No rest mass and no charge; it still carries momentum

\frac{h}{λ}

Force of a 3 W beam fully absorbed by a surface?

F

= \frac{P}{c}

= \frac{3}{3 \times 1 0 ^{8}}

= 1 0^{- 8} N

(double if reflected).

7. Standard Problem Types

7.1 What changes when you change something

Change (other things fixed)	$i_{s a t}$	$K_{ma x}$ , $V_{0}$	$ν_{0}$ , $ϕ_{0}$
Intensity doubled	doubles	same	same
Point source moved to twice the distance	$\frac{1}{4}$ (intensity $\propto \frac{1}{r ^{2}}$ )	same	same
Frequency increased, same intensity	decreases (fewer photons, $N = \frac{P}{h ν}$ )	increases	same
Frequency increased, same photon flux	same	increases	same
Metal with larger $ϕ_{0}$	depends	decreases by $Δ ϕ_{0}$	increases
Collector made more positive (after saturation)	same	same	same

7.2 Two wavelengths on one metal

Write Einstein's equation twice and subtract to remove $ϕ_{0}$ :

e (V_{1} - V_{2}) = h c (\frac{1}{λ _{1}} - \frac{1}{λ _{2}})

Exam Trick

Let $\frac{h c}{λ} = x$ and solve like algebra. " $V_{0}$ is $3 V$ for $λ$ and $V$ for $2 λ$ ": $3 e V = x - ϕ_{0}$ and $e V = \frac{x}{2} - ϕ_{0}$ . Subtract: $e V = \frac{x}{4}$ , so $ϕ_{0} = \frac{x}{4} = \frac{h c}{4 λ}$ , and $λ_{0} = 4 λ$ . No constants needed.

7.3 Other favourites

Mixed light: $V_{0}$ is set by the highest frequency present (the fastest electrons must be stopped).
Magnetic field: the fastest photoelectrons move in the largest circle, $r_{ma x}$ $= \frac{m v _{ma x}}{e B}$ $= \frac{2 m K _{ma x}}{e B}$ .
Electrons accelerated after emission: an extra accelerating potential $V$ adds $e V$ : the electrons reach A with $K_{ma x} + e V$ .
X-ray cut-off (inverse process): electrons accelerated through $V$ produce X-rays with $λ_{min} = \frac{h c}{e V} = \frac{1240}{V} nm$ ( $V$ in volts), $ν_{ma x} = \frac{e V}{h}$ .

Key idea

Remove

ϕ_{0}

by subtracting two equations;

V_{0}

follows the highest frequency present;

r_{ma x} = \frac{2 m K _{ma x}}{e B}

8. Solving Photoelectric Problems and Revision Map

Use the flowchart to pick the steps, then the mind map to revise the whole concept.

Figure 12: Solving a photoelectric numerical. Work in eV:

K_{ma x}

in eV equals

V_{0}

in volts.

Figure 13: Mind map of this concept. Cover a branch, recall its three points, then check.

9. Solved Examples

Solved Example 1

The work function of caesium is

2.14 eV

. Find (a) its threshold frequency and (b) the wavelength of light for which the stopping potential is

0.60 V

Solution:

(a) $ν_{0}$ $= \frac{ϕ _{0}}{h}$ $= \frac{2.14 \times 1.6 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}}$ $= 5.16 \times 1 0^{14} Hz$ .

(b) $\frac{1240}{λ}$ $= ϕ_{0} + e V_{0}$ $= 2.14 + 0.60$ $= 2.74 eV$ , so $λ = \frac{1240}{2.74} nm$ .

Answer: (a) $5.16 \times 1 0^{14} Hz$ ; (b) $λ \approx 453 nm$ (blue).

Solved Example 2

A He-Ne laser emits

9.42 mW

632.8 nm

. Find (a) the energy and momentum of each photon, (b) the number of photons emitted per second, and (c) the speed at which a hydrogen atom (

1.67 \times 1 0^{- 27} kg

) would have the same momentum.

Solution:

(a) $E$ $= \frac{1240}{632.8}$ $= 1.96 eV$ $= 3.14 \times 1 0^{- 19} J$ ; $p$ $= \frac{h}{λ}$ $= \frac{6.63 \times 1 0 ^{- 34}}{632.8 \times 1 0 ^{- 9}}$ $= 1.05 \times 1 0^{- 27} kg m s^{- 1}$ .

(b) $N$ $= \frac{P}{E}$ $= \frac{9.42 \times 1 0 ^{- 3}}{3.14 \times 1 0 ^{- 19}}$ $= 3.0 \times 1 0^{16} s^{- 1}$ .

Answer: (a) $1.96 eV$ , $1.05 \times 1 0^{- 27} kg m s^{- 1}$ ; (b) $3.0 \times 1 0^{16}$ photons per second; (c) $0.63 m s^{- 1}$ .

Solved Example 3

The intensity of light falling on a photosensitive surface is doubled, keeping the frequency the same. Then
(A)

K_{ma x}

doubles
(B) the saturation current doubles and

K_{ma x}

is unchanged
(C) both the saturation current and

K_{ma x}

double
(D) the stopping potential is halved

Solution:

Answer: (B). Doubling intensity doubles the number of photons per second, so twice as many electrons are emitted. Each photon still has the same $h ν$ , so $K_{ma x} = h ν - ϕ_{0}$ and $V_{0}$ do not change.

Solved Example 4

The threshold frequency of a metal is

3.3 \times 1 0^{14} Hz

. Find the stopping potential for light of frequency

8.2 \times 1 0^{14} Hz

Solution:

$e V_{0} = h (ν - ν_{0})$ , so $V_{0}$ $= \frac{6.63 \times 1 0 ^{- 34} \times ( 8.2 - 3.3 ) \times 1 0 ^{14}}{1.6 \times 1 0 ^{- 19}}$ .

Answer: $V_{0} \approx 2.0 V$ .

Solved Example 5

Light of frequency

7.21 \times 1 0^{14} Hz

ejects electrons with a maximum speed of

6.0 \times 1 0^{5} m s^{- 1}

. Find the threshold frequency. (

m_{e} = 9.11 \times 1 0^{- 31} kg

)

Solution:

$K_{ma x}$ $= \frac{1}{2} m v^{2}$ $= \frac{1}{2} \times 9.11 \times 1 0^{- 31} \times (6.0 \times 1 0^{5})^{2}$ $= 1.64 \times 1 0^{- 19} J$ ( $1.02 eV$ ).

$ν_{0}$ $= ν - \frac{K _{ma x}}{h}$ $= 7.21 \times 1 0^{14} - \frac{1.64 \times 1 0 ^{- 19}}{6.63 \times 1 0 ^{- 34}}$ $= (7.21 - 2.47) \times 1 0^{14}$ .

Answer: $ν_{0} = 4.74 \times 1 0^{14} Hz$ .

Solved Example 6

When a metal is lit with light of wavelength

λ

, the stopping potential is

3 V

; with wavelength

2 λ

it is

V

. The threshold wavelength of the metal is
(A)

2 λ

(B)

3 λ

(C)

4 λ

(D)

6 λ

Solution:

Answer: (C). $3 e V = \frac{h c}{λ} - ϕ_{0}$ and $e V = \frac{h c}{2 λ} - ϕ_{0}$ . Subtracting, $2 e V = \frac{h c}{2 λ}$ , so $e V = \frac{h c}{4 λ}$ . Then $ϕ_{0}$ $= \frac{h c}{2 λ} - \frac{h c}{4 λ}$ $= \frac{h c}{4 λ}$ , and $λ_{0} = \frac{h c}{ϕ _{0}} = 4 λ$ .

Solved Example 7

A rubidium photocell gives these stopping potentials for mercury lines:

365.0 nm

1.28 V

;

404.7 nm

0.95 V

;

435.8 nm

0.74 V

;

546.1 nm

0.16 V

;

690.7 nm

0

. Find

h

, the threshold frequency and the work function. (

e = 1.6 \times 1 0^{- 19} C

)

Solution:

Frequencies $ν = \frac{c}{λ}$ : $8.22$ , $7.41$ , $6.88$ , $5.49$ , $4.34 \times 1 0^{14} Hz$ . The $690.7 nm$ line is below threshold, so it is left out of the line (Figure 9).

Slope from the end points: $\frac{1.28 - 0.16}{( 8.22 - 5.49 ) \times 1 0 ^{14}}$ $= 4.11 \times 1 0^{- 15} V s$ (a least-squares fit of all four gives the same).

$h$ $= e \times slope$ $= 1.6 \times 1 0^{- 19} \times 4.11 \times 1 0^{- 15}$ $= 6.57 \times 1 0^{- 34} J s$ .

Threshold: $V_{0} = 0$ at $ν_{0}$ $= 8.22 \times 1 0^{14} - \frac{1.28}{4.11 \times 1 0 ^{- 15}}$ $= 5.10 \times 1 0^{14} Hz$ ; $ϕ_{0} = h ν_{0} = 2.09 eV$ .

Answer: $h \approx 6.57 \times 1 0^{- 34} J s$ , $ν_{0} \approx 5.1 \times 1 0^{14} Hz$ , $ϕ_{0} \approx 2.1 eV$ .

Solved Example 8

Photon energies at the violet (

390 nm

), yellow-green (

550 nm

) and red (

760 nm

) ends of the spectrum are needed to choose a photocell material. Which of Cs (

2.14

), K (

2.30

), Na (

2.75

), Ca (

3.20

) and Mo (

4.17 eV

) work with each colour?

Solution:

$E = \frac{1240}{λ}$ : violet $3.18 eV$ , yellow-green $2.25 eV$ , red $1.63 eV$ . Emission needs $E \geq ϕ_{0}$ .

Violet ( $3.18 eV$ ): Cs, K, Na (Ca at $3.20 eV$ just misses).
Yellow-green ( $2.25 eV$ ): Cs only.
Red ( $1.63 eV$ ): none.

Answer: violet works with Cs, K, Na; yellow-green with Cs only; red with none, so a visible-light photocell uses caesium.

Solved Example 9

Light of wavelength

300 nm

falls on a metal of work function

2.3 eV

. The photoelectrons enter a uniform magnetic field of

5.0 \times 1 0^{- 4} T

perpendicular to their velocity. Find the radius of the largest circular path.

Solution:

$K_{ma x}$ $= \frac{1240}{300} - 2.3$ $= 4.13 - 2.3$ $= 1.83 eV$ .

$p = 2 m K_{ma x}$ $= 2 \times 9.11 \times 1 0^{- 31} \times 1.83 \times 1.6 \times 1 0^{- 19}$ $= 7.31 \times 1 0^{- 25} kg m s^{- 1}$ .

$r_{ma x}$ $= \frac{p}{e B}$ $= \frac{7.31 \times 1 0 ^{- 25}}{1.6 \times 1 0 ^{- 19} \times 5.0 \times 1 0 ^{- 4}}$ .

Answer: $r_{ma x} \approx 9.1 \times 1 0^{- 3} m$ $= 9.1 mm$ .

Solved Example 10

30 mW

laser beam falls normally on a plane mirror and is completely reflected. Find the force on the mirror. What would it be if the surface absorbed the beam?

Solution:

Each photon's momentum reverses, so $Δ p = 2 p$ per photon and $F$ $= \frac{2 P}{c}$ $= \frac{2 \times 30 \times 1 0 ^{- 3}}{3 \times 1 0 ^{8}}$ .

Answer: $2 \times 1 0^{- 10} N$ on the mirror; $1 \times 1 0^{- 10} N$ on an absorbing surface.

Solved Example 11

Light containing only the wavelengths

400 nm

and

600 nm

falls on a metal of work function

2.0 eV

. The stopping potential is
(A)

0.07 V

(B)

1.10 V

(C)

1.17 V

(D)

2.07 V

Solution:

Answer: (B). Photon energies: $\frac{1240}{400} = 3.10 eV$ and $\frac{1240}{600} = 2.07 eV$ . Both emit, but $V_{0}$ must stop the fastest electrons, from $400 nm$ : $V_{0} = 3.10 - 2.0 = 1.10 V$ . Option (A) is the $600 nm$ value; (C) wrongly adds both.

Solved Example 12

An isolated metal sphere of radius

1.0 cm

and work function

4.7 eV

is lit by ultraviolet light of wavelength

200 nm

. Find the maximum potential and charge it acquires. (

\frac{1}{4 π ε _{0}}

= 9 \times 1 0^{9} N m^{2} C^{- 2}

)

Solution:

$K_{ma x}$ $= \frac{1240}{200} - 4.7$ $= 6.2 - 4.7$ $= 1.5 eV$ . As electrons leave, the sphere becomes positive and pulls them back; emission stops when its potential equals the stopping potential, $1.5 V$ .

$Q$ $= 4 π ε_{0} R V$ $= \frac{0.01 \times 1.5}{9 \times 1 0 ^{9}}$ $= 1.7 \times 1 0^{- 12} C$ (about $1 0^{7}$ electrons lost).

Answer: $V_{ma x} = 1.5 V$ , $Q \approx 1.7 \times 1 0^{- 12} C$ .

Solved Example 13

Electrons are accelerated through

30 kV

in an X-ray tube. Find the maximum frequency and minimum wavelength of the X-rays produced.

Solution:

An electron can give at most all its kinetic energy $e V$ to one photon: $h ν_{ma x} = e V$ .

$ν_{ma x}$ $= \frac{1.6 \times 1 0 ^{- 19} \times 3 \times 1 0 ^{4}}{6.63 \times 1 0 ^{- 34}}$ $= 7.24 \times 1 0^{18} Hz$ ; $λ_{min} = \frac{1240}{30000} nm$ .

Answer: $ν_{ma x} = 7.24 \times 1 0^{18} Hz$ , $λ_{min} = 0.0413 nm$ .

Practice Questions

The stopping potential in an experiment is $1.5 V$ . Find the maximum kinetic energy of the photoelectrons.Answer: $1.5 eV = 2.4 \times 1 0^{- 19} J$
Will a metal of work function $4.2 eV$ emit electrons for $330 nm$ light?Answer: No: $E = 3.76 eV < 4.2 eV$
Light of $488 nm$ gives a stopping potential of $0.38 V$ . Find the work function.Answer: $2.16 eV$
How many photons per second does a $100 W$ sodium lamp ( $589 nm$ ) emit?Answer: $2.97 \times 1 0^{20} s^{- 1}$
The slope of a $V_{0}$ - $ν$ graph is $4.12 \times 1 0^{- 15} V s$ . Find $h$ .Answer: $6.59 \times 1 0^{- 34} J s$
$227.1 nm$ light gives $V_{0} = 1.3 V$ on molybdenum. Find $ϕ_{0}$ . Will a very intense $632.8 nm$ laser eject electrons?Answer: $4.16 eV$ ; no, $1.96 eV < ϕ_{0}$ at any intensity
A photocell gives $V_{0} = 0.54 V$ for $640.2 nm$ . Predict $V_{0}$ for $427.2 nm$ .Answer: $1.51 V$
$330 nm$ light falls on Na ( $2.75$ ), K ( $2.30$ ), Mo ( $4.17$ ) and Ni ( $5.15 eV$ ). Which do not emit? What changes if the source is moved closer?Answer: Mo and Ni; moving closer only increases the current from Na and K

Common Mistakes to Avoid

Watch out

Thinking brighter light gives faster electrons. Intensity changes only the number of electrons; $K_{ma x}$ depends on frequency.
Applying $K_{ma x} = h ν - ϕ_{0}$ to every electron. It gives the maximum; most electrons come out slower.
Mixing units: $ϕ_{0}$ in eV with $h ν$ in joules. Convert with $1 eV = 1.6 \times 1 0^{- 19} J$ , or work fully in eV.
Reversing the threshold condition. Emission needs $ν \geq ν_{0}$ , that is $λ \leq λ_{0}$ (shorter wavelengths).
Using the sign of $V_{0}$ in $e V_{0} = K_{ma x}$ . Use its magnitude; the minus sign only says the collector is negative.
Assuming the photocurrent is zero at $V = 0$ . It is not; it becomes zero only at $- V_{0}$ .
Adding energies of two wavelengths in mixed light. $V_{0}$ is set by the highest frequency alone.
Believing photons are deflected by electric or magnetic fields or have rest mass. They are neutral and massless, but carry momentum $\frac{h}{λ}$ .

Frequently Asked Questions

What is the photoelectric effect?

It is the emission of electrons from a metal surface when light of frequency above the threshold frequency falls on it. The emitted electrons are called photoelectrons. Einstein explained it by treating light as photons of energy $h ν$ , each absorbed by a single electron.

What are work function and threshold frequency?

Work function is the minimum energy an electron needs to escape from a metal surface, for example $2.14 eV$ for caesium. Threshold frequency is the smallest light frequency that can cause emission, $ν_{0} = \frac{ϕ _{0}}{h}$ . Below it no electrons come out, however bright the light.

Why does maximum kinetic energy not depend on intensity?

Each electron absorbs a single photon, whose energy $h ν$ depends only on frequency. Brighter light means more photons, so more electrons are emitted, but each still gets the same $h ν$ . So $K_{ma x} = h ν - ϕ_{0}$ and the stopping potential stay unchanged.

What is stopping potential?

It is the minimum negative potential of the collector, relative to the emitter, that just stops the fastest photoelectrons, making the photocurrent zero. It measures their maximum kinetic energy: $e V_{0} = K_{ma x}$ . It depends on frequency and metal, not on intensity.

Why does the photocurrent become saturated?

As the collector is made more positive it attracts more of the emitted electrons. Once every emitted electron reaches the collector, raising the potential further cannot increase the current. This saturation current depends only on how many electrons are emitted per second, which is set by the intensity.

Why do photoelectrons have a range of kinetic energies?

Free electrons in a metal have different energies, and electrons deeper inside lose energy before reaching the surface. Only the least tightly bound electron escapes with $h ν - ϕ_{0}$ ; the others come out slower. So energies range from zero up to $K_{ma x}$ .

Which photoelectric effect questions come in NEET?

NEET often asks graph-based questions (photocurrent against intensity or potential, stopping potential against frequency), the effect of doubling intensity or frequency, threshold wavelength, and simple numericals on maximum kinetic energy and stopping potential. Working in electronvolts with 1240 divided by the wavelength in nanometres saves time.

How is the photoelectric effect tested in JEE Main and Advanced?

JEE Main asks two-wavelength problems that remove the work function, stopping potential from graphs, photon counting and saturation current. JEE Advanced adds photoelectrons in magnetic fields, radiation pressure at oblique incidence, charging of an isolated emitter and combined problems with Bohr's model.

Previous year questions on Introduction and Photoelectric Effect

29 questions from past papers, each with a step-by-step solution.

Show all 29 questions

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