NUCLEAR PHYSICS

CLASS 11,12 PHYSICS/ Physical Chemistry 

JEE MAIN REVISION

Photoelectric Effect: Complete Notes

Concepts • Einstein's Equation • Graphs • Numericals • MCQs

Understand the topic, revise the formulas and practise exam-level questions.

📚 What will you learn?

• Meaning and experimental observations of the photoelectric effect
• Work function, threshold frequency and stopping potential
• Einstein's photoelectric equation
• Important graphs and their interpretations
• Quick numerical tricks for JEE Main
• Practice MCQs with solutions

1. What is the Photoelectric Effect?

The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a material when electromagnetic radiation of sufficiently high frequency falls on it. The emitted electrons are called photoelectrons.

💡 Simple example

Imagine shining ultraviolet light on a suitable metal surface. If the light frequency is high enough, electrons may escape from the surface. This emission of electrons due to incident light is called the photoelectric effect.

Important: Light of any intensity cannot eject electrons if its frequency is below the threshold frequency of the material.

2. Important Terms

① Photoelectrons

Electrons emitted from a material due to incident electromagnetic radiation.

② Work Function (φ)

The minimum energy required to remove an electron from the surface of a material.

φ = hν₀

Here, h is Planck's constant and ν₀ is the threshold frequency.

③ Threshold Frequency (ν₀)

The minimum frequency of incident radiation required to produce photoelectric emission from a given material.

ν₀ = φ / h

④ Threshold Wavelength (λ₀)

The maximum wavelength that can produce photoelectric emission for a given material.

λ₀ = c / ν₀ = hc / φ

Emission occurs for λ ≤ λ₀, or equivalently ν ≥ ν₀.

⑤ Stopping Potential (Vₛ)

The minimum retarding potential needed to stop even the most energetic photoelectrons from reaching the collector.

K_max = eVₛ

⑥ Saturation Current

The maximum photoelectric current obtained when essentially all emitted photoelectrons are collected. It depends on the rate of photoelectron emission and the collection conditions.

3. Experimental Observations

Observation 1: Threshold frequency exists

For a given material, photoelectric emission does not occur when the incident frequency is below the threshold frequency, regardless of the light intensity in the ordinary single-photon photoelectric experiment.

Observation 2: Photoelectric current depends on intensity

For a fixed frequency above threshold and suitable collection conditions, increasing light intensity increases the number of emitted electrons per second and hence increases the photocurrent.

Observation 3: Maximum kinetic energy depends on frequency

At fixed frequency above threshold, increasing the intensity does not increase the maximum kinetic energy of photoelectrons.

Observation 4: Emission is practically instantaneous

When suitable radiation falls on a clean photosensitive surface, photoelectrons are emitted without a measurable classical energy-accumulation delay under standard experimental conditions.

4. Einstein's Photoelectric Equation

Einstein explained the photoelectric effect using the quantum nature of light. Light consists of photons, each carrying energy proportional to its frequency.

E = hν = hc / λ

When a photon interacts with an electron, its energy is used to overcome the work function. Any remaining energy becomes the kinetic energy of the emitted electron.

⭐ Most Important Formula

hν = φ + K_max

Therefore,

K_max = hν − φ

K_max = h(ν − ν₀)

K_max = hc/λ − φ

eVₛ = hν − φ

These equations are fundamental for Class 12 board examinations and JEE Main numerical questions.

5. Important Constants and Shortcuts

Quantity	Value
Planck's constant, h	6.626 × 10⁻³⁴ J s
Speed of light, c	3 × 10⁸ m/s
Electronic charge, e	1.602 × 10⁻¹⁹ C
Planck's constant in eV units	4.136 × 10⁻¹⁵ eV s
Useful product, hc	Approximately 1240 eV nm

⚡ JEE Main shortcut

Photon energy (eV) = 1240 / λ (nm)

If the wavelength is given in nanometres, calculate photon energy directly in electronvolts.

Example: For λ = 400 nm,

E = 1240 / 400 = 3.10 eV

If the work function is 2.0 eV, then K_max = 3.10 − 2.0 = 1.10 eV.

6. Important Graphs for JEE Main

Graph 1: Maximum kinetic energy vs frequency

From Einstein's equation, K_max = hν − φ.

• The graph is a straight line for a given material.
• Its slope is Planck's constant h when kinetic energy is measured in joules.
• The frequency-axis intercept is the threshold frequency ν₀.

Schematic graph; not drawn to scale.

Graph 2: Stopping potential vs frequency

Vₛ = (h/e)ν − φ/e

• Straight-line graph.
• Slope = h/e.
• Frequency-axis intercept = ν₀.

Graph 3: Photocurrent vs applied potential

• Increasing positive collector potential generally increases photocurrent until saturation.
• A reverse potential reduces photocurrent.
• At the stopping potential, the photocurrent due to photoelectrons becomes zero.
• At the same frequency, greater intensity generally produces greater saturation current.

7. Effect of Intensity and Frequency

Change	Effect on photocurrent	Effect on Kmax
Increase intensity, fixed ν above threshold	Generally increases	No change
Increase frequency, fixed intensity above threshold	Not determined by frequency alone	Increases
Increase frequency below to above threshold	Emission starts once threshold is crossed	Becomes positive above threshold
Increase positive collector potential	Increases until saturation	Maximum emitted energy remains unchanged

Remember for JEE: Intensity mainly controls the number of incident photons per second at fixed frequency. Frequency controls the energy of each photon. Do not confuse photocurrent with maximum kinetic energy.

8. Solved Numerical Examples

Example 1: Calculate maximum kinetic energy

Light of wavelength 400 nm falls on a metal whose work function is 2.0 eV. Find the maximum kinetic energy.

Show solution

Photon energy = 1240/400 = 3.10 eV.

K_max = photon energy − work function

K_max = 3.10 − 2.0 = 1.10 eV.

Example 2: Find stopping potential

The maximum kinetic energy of photoelectrons is 2.5 eV. Calculate the stopping potential.

Show solution

K_max = eVₛ. When kinetic energy is expressed in eV, its numerical value equals the stopping potential in volts.

Vₛ = 2.5 V.

Example 3: Threshold wavelength

A metal has a work function of 2.48 eV. Find its threshold wavelength.

Show solution

λ₀ = hc/φ = 1240/2.48 nm.

λ₀ = 500 nm.

Photoelectric emission is possible for wavelengths of 500 nm or less.

9. Practice Questions: JEE Main Level

Try these questions before opening the answer sections. Each question has one correct option.

Q1. The work function of a metal is 2 eV. Light of photon energy 5 eV falls on it. The maximum kinetic energy is:

1. 2 eV
2. 3 eV
3. 5 eV
4. 7 eV

Reveal answer and explanation

Answer: B — 3 eV.

K_max = 5 − 2 = 3 eV.

Q2. If the intensity of incident light is doubled while its frequency remains above threshold, the maximum kinetic energy of photoelectrons:

1. Doubles
2. Becomes half
3. Remains unchanged
4. Becomes zero

Reveal answer and explanation

Answer: C — Remains unchanged.

Maximum kinetic energy depends on frequency, not intensity, for ordinary single-photon photoelectric emission.

Q3. The work function of a metal is 3 eV. Its threshold wavelength is approximately:

1. 207 nm
2. 413 nm
3. 620 nm
4. 1240 nm

Reveal answer and explanation

Answer: B — Approximately 413 nm.

λ₀ = 1240/3 nm ≈ 413 nm.

Q4. The slope of a graph of stopping potential Vₛ against frequency ν is:

1. h
2. e/h
3. h/e
4. φ/e

Reveal answer and explanation

Answer: C — h/e.

From Vₛ = (h/e)ν − φ/e, the coefficient of frequency is h/e.

Q5. The maximum kinetic energy of photoelectrons is 1.5 eV. The stopping potential is:

1. 0.75 V
2. 1.5 V
3. 3 V
4. 1.5 eV

Reveal answer and explanation

Answer: B — 1.5 V.

K_max = eVₛ. Numerically, kinetic energy in eV equals stopping potential in volts.

Q6. For incident light with frequency below the threshold frequency, ordinary single-photon photoelectric emission:

1. Occurs if intensity is high enough
2. Cannot occur by increasing intensity alone
3. Always produces zero stopping potential with a current
4. Occurs only at high collector voltage

Reveal answer and explanation

Answer: B — Increasing intensity alone cannot produce emission.

Each photon has insufficient energy to overcome the work function.

Q7. The work function of a metal is 2.2 eV. Light of wavelength 310 nm is incident on it. The maximum kinetic energy is approximately:

1. 1.8 eV
2. 2.2 eV
3. 4.0 eV
4. 6.2 eV

Reveal answer and explanation

Answer: A — 1.8 eV.

Photon energy = 1240/310 = 4.0 eV. Therefore K_max = 4.0 − 2.2 = 1.8 eV.

Q8. In a graph of maximum kinetic energy versus frequency, the frequency-axis intercept represents:

1. The saturation frequency
2. The threshold frequency
3. The frequency of maximum current
4. The frequency of maximum intensity

Reveal answer and explanation

Answer: B — The threshold frequency.

At the intercept, K_max = 0, so ν = ν₀.

Q9. Two beams have the same frequency above threshold. Beam B has twice the intensity of beam A. Under the same collection conditions, which statement is generally correct?

1. Beam B produces twice the maximum kinetic energy
2. Beam B produces half the stopping potential
3. Beam B produces a greater saturation current
4. Beam B has half the photon energy

Reveal answer and explanation

Answer: C — Beam B produces a greater saturation current.

At fixed frequency, greater intensity means more incident photons per second and generally more emitted electrons per second.

Q10. A metal has threshold frequency 5 × 10¹⁴ Hz. Light of frequency 8 × 10¹⁴ Hz falls on it. Taking h = 4 × 10⁻¹⁵ eV s, the maximum kinetic energy is:

1. 0.8 eV
2. 1.2 eV
3. 2.0 eV
4. 3.2 eV

Reveal answer and explanation

Answer: B — 1.2 eV.

K_max = h(ν − ν₀) = 4 × 10⁻¹⁵ × 3 × 10¹⁴ eV = 1.2 eV.

10. Common Mistakes to Avoid

• Mistake: Assuming greater intensity always means greater maximum kinetic energy.
  Correct: At fixed frequency, maximum kinetic energy remains unchanged.
• Mistake: Confusing threshold wavelength with threshold frequency.
  Correct: Threshold wavelength is the maximum wavelength that can cause emission.
• Mistake: Writing K_max = hν + φ.
  Correct: K_max = hν − φ.
• Mistake: Confusing stopping potential with saturation potential.
  Correct: Stopping potential prevents photoelectrons from reaching the collector; saturation is the plateau of photocurrent.
• Mistake: Using wavelength in metres with hc = 1240 eV nm.
  Correct: Use wavelength in nanometres for that shortcut.

11. Quick Revision Sheet

Photon energy E = hν = hc/λ

Work function φ = hν₀

Threshold wavelength λ₀ = hc/φ

Einstein equation K_max = hν − φ

Stopping potential K_max = eVₛ

Wavelength shortcut E (eV) ≈ 1240/λ (nm)

Kinetic energy graph Slope = h

Stopping potential graph Slope = h/e

12. Frequently Asked Questions

Can increasing light intensity cause emission below threshold frequency?

No, not in the ordinary single-photon photoelectric effect. The energy of each photon remains below the work function.

Does the photoelectric effect prove the particle nature of light?

It provides important evidence for the quantum or particle nature of light, particularly through the photon-energy relationship.

What is the difference between work function and kinetic energy?

Work function is the minimum energy needed to remove an electron; maximum kinetic energy is the energy left over after overcoming the work function.

Which formula is most important for JEE Main?

Einstein's equation, K_max = hν − φ, together with K_max = eVₛ, is essential for many numerical problems.

Keep Practising!

Revise the concepts, remember the formulas and solve numerical problems regularly.

Class 11 Physical Chemistry| Modern Physics | JEE Main 

Educational notes for revision. Use your prescribed textbook and the latest official examination syllabus for final preparation.