Question
(JEE Main 2024 Shift 1 pattern) Light of wavelength nm falls on a metal of work function eV. Find (a) the maximum kinetic energy of the emitted photoelectrons, (b) the stopping potential, (c) the threshold wavelength for this metal. Use eV·nm.
Solution — Step by Step
Using eV·nm with in nm gives the answer directly in eV — much faster than SI units for this chapter.
Stopping potential is the voltage that stops even the fastest electron: . Since eV, we directly get:
The trick: in eV units, the numerical value of in eV equals in volts.
At threshold, photon energy equals work function exactly: .
Final Answer: (a) eV. (b) V. (c) nm.
Why This Works
Einstein’s equation says one photon ejects one electron, with the photon’s energy split between overcoming the work function and giving the electron kinetic energy. There is no time-delay or accumulation — either the photon has enough energy () or it doesn’t.
The threshold wavelength is the longest wavelength that can still eject electrons. Anything longer than nm here (e.g., red light at nm) produces no photoelectrons regardless of intensity — a fact that classical wave theory could not explain and which earned Einstein the Nobel Prize.
Alternative Method
Using SI units: J. Convert to eV by dividing by : eV. Same answer, three times more arithmetic. Stick with eV·nm in this chapter.
Plugging wavelength in metres but in eV·nm (or vice versa) is the most common slip. Either use SI throughout, or stay entirely in eV-nm. Mixing units gives a error.
Stopping potential depends only on the photon’s frequency (and the metal’s work function), not on light intensity. Intensity changes only the number of photoelectrons, not their maximum energy. This is the conceptual question NEET asks every year.