MixedJEE Physics · Original learning card5 original chapter questions

Einstein Photoelectric Equation

Energy conservation for one-photon photoemission gives h nu = phi + K_max. The stopping potential is the reverse potential whose electric potential energy e V_s equals K_max.

Why this shows up in the exam

Finding photoelectron kinetic energy · Determining stopping potential · Comparing two incident wavelengths on one metal

Learn the idea

A photon spends the work function first, and the remainder becomes electron kinetic energy. Treat photon energy like a fixed budget. The metal requires an entry cost called the work function; only the energy left after paying that cost can appear as the fastest electron's kinetic energy.

🧠 Memory hook: Photon budget equals escape cost plus maximum kinetic energy.

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

Formulas & facts to keep ready

  • h nu = phi + K_max — Einstein photoelectric equation for the fastest emitted electrons
  • K_max = e V_s = (1/2) m_e v_max² — equivalent forms of maximum photoelectron kinetic energy

How to approach it

  1. 1Write h c / lambda = phi + K_max
  2. 2Replace K_max by e V_s or one-half m v squared as needed
  3. 3Subtract two equations when the same metal is used twice

Common slip-ups that cost marks

  • •Using incident intensity in the energy equation
  • •Writing e V_s equal to photon energy without subtracting work function
  • •Mixing electron-volts and joules within one calculation

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

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

Photons of energy 5 eV illuminate a metal of work function 2 eV. Find the stopping potential.

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