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Photoelectric effect: Class 12 practice

A photon has energy E = hf = hc/λ. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Units: eV and J

The idea behind photoelectric effect

A photon has energy E = hf = hc/λ. Einstein's equation: K_max = hf − φ, where φ is the work function. No electrons are emitted below the threshold frequency f₀ = φ/h. Stopping potential V₀ = K_max/e.

Step-by-step method

  1. Find photon energy in eV (E = 1240/λ(nm)).
  2. K_max = E − φ.
  3. If K_max < 0, no emission; V₀ = K_max in volts.

Worked example 1 (Basic)

Find the energy of a photon of wavelength 250 nm in eV.

  1. E = 1240/250
  2. = 4.96 eV
  3. Answer: 4.96 eV

Worked example 2 (Standard)

Light of 350 nm falls on a metal with work function 2.1 eV. Find the maximum kinetic energy of the electrons.

  1. E = 3.54 eV
  2. K = 3.54 − 2.1 = 1.44 eV
  3. Answer: 1.44 eV

Worked example 3 (Challenge)

Find the threshold wavelength for a metal of work function 2.75 eV.

  1. λ₀ = 1240/2.75
  2. = 450.91 nm
  3. Answer: 450.91 nm

Common mistakes and misconceptions

  • Thinking brighter light raises K_max
  • Mixing J and eV

Practice questions

  1. (Basic) Find the energy of a photon of wavelength 400 nm in eV.
  2. (Basic) Find the energy of a photon of wavelength 300 nm in eV.
  3. (Standard) Light of 300 nm falls on a metal with work function 4.5 eV. Find the maximum kinetic energy of the electrons.
  4. (Standard) Light of 450 nm falls on a metal with work function 4.5 eV. Find the maximum kinetic energy of the electrons.
  5. (Challenge) Find the threshold wavelength for a metal of work function 4.5 eV.
  6. (Challenge) Find the threshold wavelength for a metal of work function 2.3 eV.

Answer key with full working

  • Q1: E = 1240/400 → = 3.1 eV → Answer: 3.1 eV
  • Q2: E = 1240/300 → = 4.13 eV → Answer: 4.13 eV
  • Q3: E = 4.13 eV → E < φ → no electrons emitted → Answer: No emission
  • Q4: E = 2.76 eV → E < φ → no electrons emitted → Answer: No emission
  • Q5: λ₀ = 1240/4.5 → = 275.56 nm → Answer: 275.56 nm
  • Q6: λ₀ = 1240/2.3 → = 539.13 nm → Answer: 539.13 nm

Syllabus note

This skill is listed under "Dual Nature of Radiation and Matter" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is photoelectric effect in?

"Dual Nature of Radiation and Matter", Class 12 Physics (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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