Einstein’s Photoelectric Equation: Derivation and explanation of photoelectric characteristics

📘 Topic Description

Einstein's Photoelectric Equation explains the photoelectric effect using the quantum theory of light. According to Einstein, light consists of discrete packets of energy called photons.

E = hν

When a photon strikes a metal surface, its energy is used partly to remove the electron from the metal surface and the remaining energy appears as the maximum kinetic energy of the emitted electron.

hν = φ + K_max

⭐ This is the most important equation of the Photoelectric Effect chapter.

🔬 Einstein's Photon Theory

① Quantization

Light energy exists in discrete packets called photons.

② Photon Energy

E = hν

③ Photon Interaction

A photon transfers its energy to an electron during interaction.

④ Instantaneous Emission

If photon energy is sufficient, emission occurs practically instantaneously.

🧮 Einstein's Photoelectric Equation — Derivation

Energy of incident photon:

E = hν

Minimum energy required to remove an electron:

φ = hν₀

Remaining energy becomes maximum kinetic energy:

hν = φ + K_max

Therefore:

K_max = hν − φ

Since φ = hν₀:

⭐ K_max = h(ν − ν₀)

Also:

K_max = ½mv_max²

Hence:

⭐ hν = φ + ½mv_max²

⚡ Stopping Potential

If V₀ is the stopping potential, then:

K_max = eV₀

hν = φ + eV₀

V₀ = (h/e)ν − φ/e

Stopping potential is directly related to the maximum kinetic energy of photoelectrons.

📈 Photoelectric Characteristics

Frequency Effect

Maximum kinetic energy increases with frequency.

K_max = hν − φ

Intensity Effect

For ν > ν₀, increasing intensity increases the number of photoelectrons and hence photoelectric current.

Threshold Frequency

No photoelectric emission occurs when ν < ν₀.

Instantaneous Emission

Emission occurs practically immediately when ν ≥ ν₀.

📊 Interactive K_max vs Frequency

Frequency ν

Work Function φ

🔋 Work Function

The work function is the minimum energy required to remove an electron from a metal surface.

φ = hν₀

SI unit = Joule (J)
Common unit = electron volt (eV)

🚫 Threshold Frequency

ν₀ = φ/h

ν < ν₀ → No emission
ν = ν₀ → K_max = 0
ν > ν₀ → Photoelectric emission

💡 Effect of Intensity

• Increasing intensity increases the number of photons per second.
• More photons eject more electrons.
• Photoelectric current increases.
• Maximum kinetic energy does not increase due to intensity alone.

Intensity ↑ → Photoelectric Current ↑

📡 Effect of Frequency

• Increasing frequency increases photon energy.
• Maximum kinetic energy increases.
• Stopping potential increases.
• Threshold frequency depends on the metal.

ν ↑ → K_max ↑ → V₀ ↑

📈 Important Photoelectric Graphs

Graph	Nature	Important Result
Kmax vs ν	Straight Line	Slope = h
V0 vs ν	Straight Line	Slope = h/e
Photocurrent vs Intensity	Increasing	Current increases
Photocurrent vs Voltage	Saturation Curve	Saturation current

🧠 Why Classical Wave Theory Failed?

• It could not explain threshold frequency.
• It could not explain instantaneous emission.
• It could not explain frequency dependence of maximum kinetic energy.
• It could not correctly explain the effect of intensity.

Einstein explained these observations by treating light as photons.

🔥 Important Search Keywords

Einstein Photoelectric Equation Photoelectric Effect Class 12 Photoelectric Equation Derivation Photoelectric Effect Numericals Photoelectric Effect MCQ Work Function Threshold Frequency Stopping Potential Maximum Kinetic Energy Photon Energy Dual Nature of Radiation CBSE Class 12 Physics JEE Physics NEET Physics Modern Physics

📝 MCQ Practice — 15 Questions

1. Einstein's photoelectric equation is:

A. hν = φ − K

B. hν = φ + K_max

C. hν = φK

D. hν = K − φ

✔ Answer: B

2. Photon energy is:

A. h/ν

B. hν

C. ν/h

D. hcν

✔ Answer: B

3. Work function is equal to:

A. hν₀

B. h/ν₀

C. ν₀/h

D. hc/ν₀

✔ Answer: A

4. Maximum kinetic energy mainly depends on:

A. Intensity

B. Frequency

C. Area

D. Distance

✔ Answer: B

5. Increasing intensity above threshold frequency increases:

A. K_max

B. Work function

C. Photoelectric current

D. Threshold frequency

✔ Answer: C

6. If ν = ν₀, then:

A. K_max = 0

B. K_max is maximum

C. Current is infinite

D. Photon energy is zero

✔ Answer: A

7. The slope of K_max versus ν graph is:

A. e

B. h

C. 1/h

D. φ

✔ Answer: B

8. Maximum kinetic energy is related to stopping potential by:

A. K_max = eV₀

B. K_max = V₀/e

C. K_max = hV₀

D. K_max = φV₀

✔ Answer: A

9. Threshold frequency depends upon:

A. Intensity

B. Nature of metal

C. Distance

D. Area

✔ Answer: B

10. Photoelectric emission is:

A. Delayed

B. Practically instantaneous

C. Always impossible

D. Periodic

✔ Answer: B

11. If frequency increases, photon energy:

A. Decreases

B. Increases

C. Remains same

D. Becomes zero

✔ Answer: B

12. Which equation represents work function?

A. φ = hν₀

B. φ = h/ν₀

C. φ = eV₀

D. φ = mc²

✔ Answer: A

13. If frequency is below threshold frequency:

A. Slow emission occurs

B. Fast emission occurs

C. No photoelectric emission occurs

D. Kinetic energy becomes infinite

✔ Answer: C

14. Einstein's explanation supports the:

A. Particle nature of light

B. Sound nature of light

C. Mechanical nature of light

D. Thermal nature only

✔ Answer: A

15. Maximum kinetic energy is:

A. h(ν + ν₀)

B. h(ν − ν₀)

C. hνν₀

D. h/(ν − ν₀)

✔ Answer: B

🟣 Assertion–Reason — 5 Questions

A. Both A and R are true and R is the correct explanation of A.
B. Both A and R are true but R is not the correct explanation of A.
C. A is true but R is false.
D. A is false but R is true.

1. Assertion: Maximum kinetic energy increases with frequency.

Reason: Photon energy is E = hν.

Answer: A

2. Assertion: Increasing intensity increases maximum kinetic energy.

Reason: Photon energy depends on frequency.

Answer: D

3. Assertion: No photoelectric emission occurs below threshold frequency.

Reason: Photon energy is less than work function.

Answer: A

4. Assertion: Stopping potential increases with frequency.

Reason: Maximum kinetic energy increases with frequency.

Answer: A

5. Assertion: Work function is independent of metal.

Reason: Work function is a characteristic property of a metal.

Answer: D

🟢 2 Marks — 6 Questions

Q1. State Einstein's photoelectric equation.

Q2. Define work function.

Q3. Define threshold frequency.

Q4. Write the relation between stopping potential and maximum kinetic energy.

Q5. Why does maximum kinetic energy depend on frequency?

Q6. Why does intensity not affect maximum kinetic energy?

🟡 3 Marks — 6 Questions

Q1. Derive φ = hν₀.

Q2. Explain the physical meaning of Einstein's equation.

Q3. Explain the effect of intensity on photoelectric current.

Q4. Explain the effect of frequency on K_max.

Q5. Explain stopping potential.

Q6. Why does photoelectric emission occur instantaneously?

🟠 4 Marks — 6 Questions

Q1. Derive Einstein's photoelectric equation.

Q2. Explain threshold frequency and work function.

Q3. Explain frequency dependence of maximum kinetic energy.

Q4. Explain intensity dependence of photoelectric current.

Q5. Explain the K_max versus frequency graph.

Q6. Explain stopping potential versus frequency.

🔴 5 Marks — 6 Questions

Q1. Derive Einstein's photoelectric equation and explain every term.

Q2. Explain major photoelectric characteristics using Einstein's theory.

Q3. Derive V₀ = (h/e)ν − φ/e.

Q4. Explain why classical wave theory cannot explain photoelectric effect.

Q5. Explain particle nature of light using photoelectric effect.

Q6. Explain K_max versus frequency and V₀ versus frequency graphs.

🔵 6 Marks — 6 Questions

Q1. Give a complete derivation of Einstein's photoelectric equation.

Q2. Explain experimental characteristics of photoelectric effect using Einstein's theory.

Q3. Derive stopping potential equation and explain its graph.

Q4. Explain how photoelectric effect proves quantum nature of electromagnetic radiation.

Q5. Explain the role of work function and threshold frequency.

Q6. Discuss effects of intensity and frequency on photocurrent and kinetic energy.

🧮 Numerical Practice — 6 Questions

Q1. A metal has work function 2 eV. Calculate its threshold frequency. Use h = 4.14 × 10⁻¹⁵ eV s.

Q2. Light of frequency 8 × 10¹⁴ Hz falls on a metal having threshold frequency 5 × 10¹⁴ Hz. Find maximum kinetic energy.

Q3. The work function of a metal is 3 eV. Calculate threshold frequency.

Q4. Radiation of frequency 1 × 10¹⁵ Hz falls on a metal whose work function is 2 eV. Find maximum kinetic energy.

Q5. A photoelectron has maximum kinetic energy 2 eV. Find stopping potential.

Q6. The threshold wavelength of a metal is 400 nm. Calculate threshold frequency.

📌 Complete Formula Revision

Photon Energy

E = hν

Einstein Equation

hν = φ + K_max

Maximum Kinetic Energy

K_max = hν − φ

Threshold Frequency

ν₀ = φ/h

Stopping Potential

K_max = eV₀

Combined Equation

hν = φ + eV₀

Maximum Velocity

hν = φ + ½mv_max²

Threshold Wavelength

λ₀ = c/ν₀

🚀 One-Minute Revision

🔹 Photon energy = hν
🔹 Work function = hν₀
🔹 Einstein equation = hν = φ + K_max
🔹 K_max = h(ν − ν₀)
🔹 K_max = eV₀
🔹 ν < ν₀ → No emission
🔹 ν = ν₀ → K_max = 0
🔹 ν > ν₀ → Photoelectric emission
🔹 Intensity ↑ → Photocurrent ↑
🔹 Frequency ↑ → K_max ↑
🔹 Slope of K_max vs ν = h
🔹 Slope of V₀ vs ν = h/e

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⚡ Einstein's Photoelectric Equation • Complete Practice Material