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Photoelectric Effect & Einstein's Explanation

Master the Photoelectric Effect for JEE Main, JEE Advanced, NEET & Board Exams. Covers Hertz and Lenard observations, Einstein photon theory, photoelectric equation, work function, stopping potential, threshold frequency, wave-particle duality, Millikan experiment and radiation pressure with solved examples.

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The Photoelectric Effect is the phenomenon of emission of electrons from a metal surface when light of sufficiently high frequency falls on it. First observed by Heinrich Hertz in 1887 and experimentally studied in detail by Philipp Lenard, it posed a fundamental challenge to classical wave theory — the experimental observations simply could not be explained by treating light as a wave. It was Albert Einstein's brilliant 1905 explanation, for which he received the Nobel Prize in Physics (1921), that revolutionised physics by establishing the quantum (photon) nature of light. This chapter marks the beginning of quantum physics and is a cornerstone of the JEE and NEET syllabus, contributing 2–3 questions every year.

1. Hertz and Lenard's Experimental Observations

When ultraviolet light falls on a freshly cleaned zinc plate connected to a gold-leaf electroscope, the electroscope discharges if the plate was negatively charged, but not if positively charged. This indicated that negatively charged particles (electrons) are emitted from the metal surface.

Lenard's Experimental Setup

Lenard used an evacuated glass tube with two electrodes — an emitter (cathode) and a collector (anode) — connected to a battery and galvanometer. When light fell on the emitter:

A current (called photoelectric current) flowed in the circuit, confirming emission of electrons.
The emitted electrons were called photoelectrons.

Key Experimental Observations

Observation	Detail	Classical Prediction	Matches Classical Theory?
Threshold frequency	Emission occurs only if $ν \geq ν_{0}$ , regardless of intensity	Any frequency should work given enough intensity	No ✗
Effect of intensity	Intensity affects number of photoelectrons (current), NOT their maximum KE	Higher intensity should give electrons more energy	No ✗
Effect of frequency	Maximum KE of photoelectrons increases linearly with frequency	KE should depend on intensity, not frequency	No ✗
Instantaneous emission	Emission is instantaneous — no time lag between light falling and emission	Energy accumulation should take time (seconds to minutes)	No ✗

All four key observations contradicted classical wave theory — demonstrating that a fundamentally new model of light was required.

2. Einstein's Photon Theory (1905)

Einstein extended Planck's quantum hypothesis to propose that light itself consists of discrete packets of energy called photons (or quanta). Each photon has energy:

$E = h ν = \frac{h c}{λ}$

where $h = 6.626 \times 10^{- 34}$ J·s is Planck's constant, $ν$ is the frequency, $c = 3 \times 10^{8}$ m/s is the speed of light, and $λ$ is the wavelength.

Properties of Photons

Property	Value / Expression
Energy	$E = h ν = h c / λ$
Rest mass	Zero ( $m_{0} = 0$ )
Speed	Always $c = 3 \times 10^{8}$ m/s (in vacuum)
Momentum	$p = h ν / c = h / λ = E / c$
Charge	Zero (electrically neutral)
Effect of increasing intensity	More photons per second (not more energetic photons)
Effect of increasing frequency	Each photon has more energy ( $E \propto ν$ )

3. Einstein's Photoelectric Equation

When a photon of energy $h ν$ strikes the metal surface, it gives all its energy to a single electron. Part of this energy is used to overcome the binding force of the metal (work function $ϕ_{0}$ ), and the rest appears as kinetic energy of the emitted electron:

$h ν = ϕ_{0} + K E_{m a x}$

$K E_{m a x} = h ν - ϕ_{0} = h (ν - ν_{0})$

This is Einstein's Photoelectric Equation — one of the most important equations in modern physics.

Key Terms

Term	Symbol	Definition	Unit
Work Function	$ϕ_{0}$ or $W$	Minimum energy required to emit an electron from metal surface	eV or Joule
Threshold Frequency	$ν_{0}$	Minimum frequency of light for photoelectric emission: $ν_{0} = ϕ_{0} / h$	Hz
Threshold Wavelength	$λ_{0}$	Maximum wavelength for emission: $λ_{0} = h c / ϕ_{0}$	m (nm)
Maximum KE	$K E_{m a x}$	KE of electrons emitted from the surface (with zero binding energy loss)	eV or Joule
Stopping Potential	$V_{0}$	Minimum retarding potential to stop all photoelectrons: $e V_{0} = K E_{m a x}$	Volt (V)

Relations Between Key Quantities

$ϕ_{0} = h ν_{0} = \frac{h c}{λ_{0}}$
$K E_{m a x} = e V_{0} = h ν - ϕ_{0} = h ν - h ν_{0} = h c (\frac{1}{λ} - \frac{1}{λ_{0}})$
$V_{0} = \frac{h}{e} (ν - ν_{0}) = \frac{h}{e} ν - \frac{ϕ_{0}}{e}$

The $V_{0}$ – $ν$ graph is a straight line with slope $h / e$ and y-intercept $- ϕ_{0} / e$ . Millikan used this to experimentally determine Planck's constant $h$ very accurately.

4. Experimental Study — Effect of Intensity and Frequency

Effect of Intensity (at fixed frequency $ν > ν_{0}$ )

Increasing intensity → more photons per second → more photoelectrons emitted per second → photoelectric current increases.
Maximum KE of photoelectrons does NOT change — each photon gives the same energy $h ν$ , and work function $ϕ_{0}$ is fixed.
Stopping potential $V_{0}$ remains the same.

Effect of Frequency (at fixed intensity)

Increasing frequency → each photon has more energy $h ν$ → more energy given to each electron → maximum KE increases.
Stopping potential $V_{0}$ increases linearly with $ν$ .
The number of photoelectrons (current) may slightly decrease since photon energy increases but number of photons decreases (fixed intensity means fixed power).

Saturation Current

When a positive voltage is applied to the collector (anode), the current reaches a maximum value called the saturation current — all emitted photoelectrons are collected. Beyond saturation voltage, current does not increase.

I–V Characteristics

Parameter Changed	Effect on Stopping Potential $V_{0}$	Effect on Saturation Current
Increase intensity (fixed $ν$ )	No change	Increases
Increase frequency (fixed intensity)	Increases	May slightly decrease
Change metal (higher $ϕ_{0}$ )	Decreases (higher $ϕ_{0}$ means less KE)	May decrease

5. Work Functions of Common Metals

Metal	Work Function $ϕ_{0}$ (eV)	Threshold Wavelength $λ_{0}$ (nm)
Caesium (Cs)	$2.14$	$\approx 580$ (visible)
Sodium (Na)	$2.75$	$\approx 451$ (visible)
Potassium (K)	$2.30$	$\approx 539$ (visible)
Zinc (Zn)	$4.31$	$\approx 288$ (UV)
Copper (Cu)	$4.65$	$\approx 267$ (UV)
Platinum (Pt)	$5.65$	$\approx 219$ (UV)

Alkali metals (Cs, Na, K) have low work functions — they show the photoelectric effect even with visible light. Most other metals require ultraviolet light.

6. Einstein's Explanation of Observations

Observation	Einstein's Explanation
Threshold frequency exists	A photon must have at least energy $ϕ_{0}$ to eject an electron. Below $ν_{0}$ , $h ν < ϕ_{0}$ — no emission possible regardless of intensity.
Intensity affects current, not KE	More intensity = more photons per second = more electrons per second (more current). Each individual photon-electron interaction is unchanged, so KE is unchanged.
Frequency increases KE linearly	$K E_{m a x} = h ν - ϕ_{0}$ . Since $ϕ_{0}$ is fixed, KE increases linearly with $ν$ . Slope = $h$ (Planck's constant).
Instantaneous emission	One photon gives all its energy to one electron instantly — no accumulation needed. The interaction is like a collision between two particles.

7. Dual Nature of Light — Wave-Particle Duality

The photoelectric effect establishes that light has a particle nature (photons). But phenomena like interference and diffraction establish the wave nature of light. Light therefore exhibits wave-particle duality:

Wave Nature Evidence	Particle Nature Evidence
Interference (Young's double slit)	Photoelectric effect
Diffraction	Compton effect (scattering of X-rays)
Polarisation	Photon momentum ( $p = h / λ$ )
Refraction, reflection	Radiation pressure

Bohr's Complementarity Principle: Light behaves as a wave or as a particle depending on the type of experiment performed — it never shows both aspects simultaneously in the same experiment.

Radiation Pressure

Since photons carry momentum $p = h / λ$ , when light falls on a surface it exerts a pressure called radiation pressure:

For perfectly absorbing surface: Radiation pressure $= \frac{I}{c}$ (where $I$ is intensity)
For perfectly reflecting surface: Radiation pressure $= \frac{2 I}{c}$

8. Millikan's Experiment — Verification of Einstein's Equation

Robert Millikan (1916) experimentally verified Einstein's photoelectric equation by measuring the stopping potential $V_{0}$ for different frequencies of light on different metals.

He plotted $V_{0}$ vs $ν$ and obtained a straight line for every metal tested.
The slope was the same for all metals: $slope = h / e = 4.14 \times 10^{- 15}$ V·s.
Using $e = 1.6 \times 10^{- 19}$ C, he got $h = 6.626 \times 10^{- 34}$ J·s — in excellent agreement with Planck's value.
The x-intercept of the $V_{0}$ – $ν$ graph gives $ν_{0}$ (threshold frequency) — different for different metals.
The y-intercept gives $- ϕ_{0} / e$ — the work function divided by electron charge.

Millikan won the Nobel Prize in Physics (1923) for this work and for measuring the charge of the electron.

9. Important Numerical Values

Constant	Symbol	Value
Planck's constant	$h$	$6.626 \times 10^{- 34}$ J·s $= 4.14 \times 10^{- 15}$ eV·s
Speed of light	$c$	$3 \times 10^{8}$ m/s
$h c$ product	$h c$	$1240$ eV·nm (extremely useful!)
Electron charge	$e$	$1.6 \times 10^{- 19}$ C
Electron mass	$m_{e}$	$9.11 \times 10^{- 31}$ kg
$h / e$	—	$4.14 \times 10^{- 15}$ V·s (slope of $V_{0}$ – $ν$ graph)
1 eV	—	$1.6 \times 10^{- 19}$ J

Golden shortcut: $E (eV) = \frac{1240}{λ (nm)}$ — this single formula converts wavelength to photon energy in eV instantly, saving enormous calculation time in JEE and NEET.

Solving Photoelectric Effect Problems — Master Strategy

Problem Type	Approach	Key Formula
Find photon energy from wavelength	Use $h c$ shortcut	$E = 1240 / λ$ (nm) eV
Find max KE of photoelectron	Subtract work function from photon energy	$K E_{m a x} = h ν - ϕ_{0}$
Find stopping potential	Convert KE in eV to volts directly	$V_{0} = K E_{m a x} (in eV)$ volts
Find threshold wavelength	Set $K E = 0$ → photon energy = work function	$λ_{0} = 1240 / ϕ_{0} (eV)$ nm
Find $h$ from $V_{0}$ – $ν$ graph	Slope = $h / e$ ; multiply slope by $e$	$h = e \times$ slope
Photon momentum	Use $p = h / λ$ (no mass involved)	$p = h / λ = E / c$
Will emission occur?	Compare photon energy with work function	If $h ν \geq ϕ_{0}$ → yes; else → no

JEE/NEET Power Tips — Avoid These Common Mistakes

Stopping potential depends on frequency, NOT intensity. This is the most tested conceptual point. Doubling the intensity doubles the current but leaves $V_{0}$ unchanged. Increasing frequency increases $V_{0}$ .
Saturation current depends on intensity, NOT frequency. More photons per second (higher intensity) means more electrons per second (more current). More energetic photons (higher frequency) don't increase the number of electrons.
$K E_{m a x}$ is the energy of electrons emitted from the surface — electrons from deeper inside the metal have less KE because they lose energy escaping. $K E_{m a x}$ represents the luckiest electron (surface electron with zero extra binding).
Use $E = 1240 / λ$ (nm) eV for instant calculation — this is the single most time-saving shortcut in the entire chapter. Memorise it.
Photons have zero rest mass but non-zero momentum. $p = h / λ$ — do not confuse this with classical momentum $m v$ . A photon's momentum is entirely due to its energy: $p = E / c$ .
Below threshold frequency, no emission occurs even if intensity is extremely high — a million dim photons cannot eject an electron if each carries insufficient energy. One-photon one-electron interaction.
Einstein won the Nobel Prize for the photoelectric effect, not relativity. This is a famous fact often tested directly in NEET.

Practice Questions (JEE / NEET Level)

Q1: Light of wavelength 400 nm falls on a metal with a work function of 1.5 eV. The maximum kinetic energy of the emitted photoelectrons is:

A) 1.1 eV
B) 1.6 eV
C) 3.1 eV
D) 0.5 eV

Answer: B) 1.6 eV.

Explanation:

First, calculate the energy of the incident photon using the shortcut formula $E = \frac{1240}{λ (in nm)}$ eV:
Photon energy $= \frac{1240}{400} = 3.1 eV$ .

Use Einstein's photoelectric equation to find the maximum kinetic energy:
$K E_{max} = h ν - ϕ_{0}$
$K E_{max} = 3.1 - 1.5 = 1.6 eV$ .

Q2: The stopping potential for a photoelectric experiment is 2.4 V. The maximum kinetic energy of photoelectrons in joules is:

A) $2.4 \times 10^{- 19}$ J
B) $3.84 \times 10^{- 19}$ J
C) $1.6 \times 10^{- 19}$ J
D) $4.8 \times 10^{- 19}$ J

Answer: B) $3.84 \times 10^{- 19}$ J.

Explanation:

The maximum kinetic energy is directly related to the stopping potential ( $V_{0}$ ) by the equation $K E_{max} = e V_{0}$ .

Substitute the elementary charge ( $e = 1.6 \times 10^{- 19}$ C) and the stopping potential:
$K E_{max} = 1.6 \times 10^{- 19} \times 2.4$
$K E_{max} = 3.84 \times 10^{- 19} J$ .

Q3: The work function of a metal is 4.2 eV. The threshold wavelength for the photoelectric effect is approximately:

A) 296 nm
B) 395 nm
C) 195 nm
D) 495 nm

Answer: A) 296 nm.

Explanation:

The threshold wavelength ( $λ_{0}$ ) is the maximum wavelength that can cause photoelectric emission. It is calculated using $λ_{0} = \frac{1240}{ϕ_{0} (in eV)} nm$ .

$λ_{0} = \frac{1240}{4.2} \approx 295.2 nm \approx 296 nm$ .

This wavelength falls in the ultraviolet region, meaning this specific metal requires UV light to emit photoelectrons.

Q4: In a photoelectric experiment, when the intensity of incident light is doubled (frequency kept constant above threshold), which of the following is correct?

A) Both stopping potential and photoelectric current double
B) Stopping potential doubles, current stays the same
C) Stopping potential stays the same, current doubles
D) Both stopping potential and current remain unchanged

Answer: C) Stopping potential stays the same, current doubles.

Explanation:

Doubling the intensity (at a constant frequency) means there are twice as many photons hitting the surface per second. Because each individual photon still carries the same amount of energy ( $h ν$ ), the maximum kinetic energy of the emitted electrons ( $K E_{max} = h ν - ϕ_{0}$ ) remains unchanged. Therefore, the stopping potential ( $V_{0}$ ) stays the same.

However, because there are twice as many incident photons, twice as many photoelectrons are emitted per second, which causes the photoelectric current to double.

Q5: In Millikan's photoelectric experiment, the stopping potential $V_{0}$ versus frequency $ν$ graph is plotted for two metals A and B. The graph for A has a higher x-intercept (threshold frequency) than B. Which statement is correct?

A) Metal A has a higher work function than B
B) Metal A has a lower work function than B
C) Both metals have the same work function
D) The slope for A is steeper than for B

Answer: A) Metal A has a higher work function than B.

Explanation:

On a $V_{0}$ versus $ν$ graph, the x-intercept represents the threshold frequency ( $ν_{0}$ ). The relationship between work function ( $ϕ_{0}$ ) and threshold frequency is $ϕ_{0} = h ν_{0}$ .

Because Metal A has a higher threshold frequency ( $ν_{0}$ ), it must inherently have a higher work function ( $ϕ_{0}$ ). Note that the slope of these graphs is always $h / e$ , which is a universal constant independent of the metal used.

Q6: A photon of wavelength 3000 Å strikes a metal surface. The work function of the metal is 4.5 eV. The maximum velocity of the emitted photoelectron is: ( $m_{e} = 9.1 \times 10^{- 31}$ kg, $h = 6.6 \times 10^{- 34}$ J·s)

A) $6.5 \times 10^{5}$ m/s
B) No emission occurs
C) $1.3 \times 10^{6}$ m/s
D) $9.8 \times 10^{5}$ m/s

Answer: B) No emission occurs.

Explanation:

First, convert the wavelength from Angstroms to nanometers: 3000 Å = 300 nm.

Calculate the energy of the incident photon:
Photon energy $= \frac{1240}{300} \approx 4.13 eV$ .

The work function of the metal is $ϕ_{0} = 4.5 eV$ . Since the incident photon energy ( $4.13 eV$ ) is strictly less than the required work function ( $4.5 eV$ ), the photon does not have enough energy to eject an electron. Therefore, no photoelectric emission occurs.

Q7 (JEE Advanced type): Light of intensity $I$ and frequency $ν$ ( $> ν_{0}$ ) falls on a photosensitive surface. The number of photoelectrons emitted per second is $N$ and the maximum kinetic energy of each is $E_{k}$ . If both the intensity and frequency are doubled, the new values of the number of photoelectrons per second ( $N^{'}$ ) and the maximum kinetic energy ( $E_{k}^{'}$ ) are:

A) $N^{'} = N$ , $E_{k}^{'} = 2 E_{k}$
B) $N^{'} = 2 N$ , $E_{k}^{'} = 2 E_{k}$
C) $N^{'} = N$ , $E_{k}^{'} = 2 h ν - ϕ_{0}$
D) $N^{'} = 2 N$ , $E_{k}^{'} = 2 h ν - ϕ_{0}$

Answer: C) $N^{'} = N$ , $E_{k}^{'} = 2 h ν - ϕ_{0}$ .

Explanation:

Kinetic Energy:
When the frequency is doubled to $2 ν$ , the new maximum kinetic energy is:
$E_{k}^{'} = h (2 ν) - ϕ_{0} = 2 h ν - ϕ_{0}$ .
(Note: This is not simply $2 E_{k}$ , because $2 E_{k} = 2 (h ν - ϕ_{0}) = 2 h ν - 2 ϕ_{0}$ .)

Number of Photoelectrons:
Intensity ( $I$ ) is defined as the total energy per unit area per unit time, which equals the number of photons per second multiplied by the energy of a single photon: $I = N_{photons} \times h ν$ .

If both intensity and frequency are doubled:
$2 I = N_{photons}^{'} \times h (2 ν)$
$N_{photons}^{'} = \frac{2 I}{2 h ν} = \frac{I}{h ν} = N_{photons}$

Since the photon flux remains exactly the same, the number of photoelectrons emitted per second also remains unchanged ( $N^{'} = N$ ).

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Photoelectric Effect & Einstein's Explanation

1. Hertz and Lenard's Experimental Observations

Lenard's Experimental Setup

Key Experimental Observations

2. Einstein's Photon Theory (1905)

Properties of Photons

3. Einstein's Photoelectric Equation

Key Terms

Relations Between Key Quantities

4. Experimental Study — Effect of Intensity and Frequency

Effect of Intensity (at fixed frequency $ν > ν_{0}$ )

Effect of Frequency (at fixed intensity)

Saturation Current

I–V Characteristics

5. Work Functions of Common Metals

6. Einstein's Explanation of Observations

7. Dual Nature of Light — Wave-Particle Duality

Radiation Pressure

8. Millikan's Experiment — Verification of Einstein's Equation

9. Important Numerical Values

Solving Photoelectric Effect Problems — Master Strategy

Practice Questions (JEE / NEET Level)

Related Study Material

Practice Questions

Topic Information

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Track Your Learning

Study Tips

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Target Subjects

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Subject Mastery

Photoelectric Effect & Einstein's Explanation

1. Hertz and Lenard's Experimental Observations

Lenard's Experimental Setup

Key Experimental Observations

2. Einstein's Photon Theory (1905)

Properties of Photons

3. Einstein's Photoelectric Equation

Key Terms

Relations Between Key Quantities

4. Experimental Study — Effect of Intensity and Frequency

Effect of Intensity (at fixed frequency ν>ν0)

Effect of Frequency (at fixed intensity)

Saturation Current

I–V Characteristics

5. Work Functions of Common Metals

6. Einstein's Explanation of Observations

7. Dual Nature of Light — Wave-Particle Duality

Radiation Pressure

8. Millikan's Experiment — Verification of Einstein's Equation

9. Important Numerical Values

Solving Photoelectric Effect Problems — Master Strategy

Practice Questions (JEE / NEET Level)

Related Study Material

Practice Questions

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Effect of Intensity (at fixed frequency $ν > ν_{0}$ )