Exam level12 past questions

Photoelectric Effect

The photoelectric effect describes the emission of electrons from a material when it is exposed to light of sufficient frequency, governed by concepts such as threshold frequency, work function, stopping potential, and the Einstein photoelectric equation.

Why this shows up in the exam

NEET tests your understanding of how light interacts with matter, including calculations involving threshold frequency, kinetic energy, stopping potential, and the interpretation of experimental data.

How NEET tests this

Numerical · 7 QsStatement analysis · 3 QsApplication · 3 Qs

Learn the idea

When light of frequency above a material's threshold strikes it, each photon can liberate one electron; the excess photon energy becomes the electron's kinetic energy (Einstein's equation). The key insight is that frequency (not intensity) decides whether electrons are emitted, while intensity controls how many are emitted.

🧠 Memory hook: Frequency opens the door (emission), intensity fills the room (current).

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Photon energy E = hν = hc/λ
  • Threshold frequency ν₀ = φ/h and threshold wavelength λ₀ = hc/φ
  • Einstein photoelectric equation KE_max = hν – φ = eV_s
  • Photoelectric current ∝ light intensity (number of photons) for ν > ν₀
  • If ν ≤ ν₀ no electrons are emitted regardless of intensity

How to approach it

  1. 11) Verify ν > ν₀ (or λ < λ₀) to confirm emission.
  2. 22) Compute photon energy using E = hc/λ (or hν).
  3. 33) Subtract the work function φ to get KE_max; if stopping potential is required, use V_s = KE_max/e.
  4. 44) For current, relate intensity to photon flux; for speed, use KE_max = ½mv².

Worked example — watch it click

The photoelectric threshold wavelength of silver is 3250 × 10⁻¹⁰ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10⁻¹⁰ m is: (Given h = 4.14×10⁻¹⁵ eVs and c = 3×10⁸ m/s)

  • A)0.6 x 10⁶ m/s
  • B)61 x 10³ m/s
  • C)0.3 x 10⁶ m/s
  • ✅6.1 x 10⁵ m/s

The concept behind this problem

The example forces you to turn a wavelength into photon energy, subtract the work function, and then translate the remaining kinetic energy into electron speed – exactly the steps of Einstein's photoelectric equation.

Step by step

  1. 1Work function φ = hc/λ₀ = (4.14×10⁻¹⁵ × 3×10⁸)/(3250×10⁻¹⁰) = 3.82 eV.
  2. 2Incident photon energy E = hc/λ = (4.14×10⁻¹⁵ × 3×10⁸)/(2536×10⁻¹⁰) = 4.90 eV.
  3. 3KE = E - φ = 4.90 - 3.82 = 1.08 eV = 1.73×10⁻¹⁹ J.
  4. 4Using ½mv² = KE: v = √(2×1.73×10⁻¹⁹/9.1×10⁻³¹) ≈ 6.2×10⁵ m/s ≈ 6.1×10⁵ m/s.

Watch out

Students often skip converting the work function from eV to joules before using KE = ½mv², giving an incorrect speed.

Common slip-ups that cost marks

  • •Mixing up intensity with kinetic energy – intensity changes current, not KE.
  • •Using λ instead of λ₀ (or ν instead of ν₀) and concluding emission when there is none.
  • •For speed calculations, forgetting to convert eV to joules before applying ½mv².

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 12

Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled?

Push further

More challenging

24 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 24

A metallic surface has a threshold frequency of 6.0 x 10¹⁴ Hz. If light of frequency 9.0 x 10¹⁴ Hz is incident on it, photoelectrons are emitted. What happens to the photoelectric current if the incident light's frequency is changed to 4.0 x 10¹⁴ Hz, while its intensity is kept constant?