JEE Main 2022 · 29 July · morning · Single correct · +4 / −1
The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength:
- (a) hc/(E λ-hc)
- (b) (hc λ)/(E λ+hc)
- (c) (h λ)/(E λ+hc)
- (d) (hc λ)/(E λ-hc)
Answer and solution
Answer (b) (hc λ)/(E λ+hc)
Solution
Physics sir · 4 steps
Write Einstein's equation for both cases and subtract, so the unknown work function drops out.
- Two equations. With work function φ, the first light gives E = hc/λ - φ. The new wavelength λ' must give twice the kinetic energy: 2E = hc/λ' - φ.
- Subtract. Take the first equation away from the second. The φ cancels and leaves E = hc/λ' - hc/λ, so the new photon carries exactly E more energy than the old one.
hc/λ' = E + hc/λ = (Eλ + hc)/λ
- Invert. Flip both sides and multiply by hc: λ' = hcλ/(Eλ + hc). The denominator is bigger than hc, so λ' < λ, which is right for a more energetic photon.
- So. The required wavelength is hcλ/(Eλ + hc), option (b).
The idea to keep. When the work function is not given, write the photoelectric equation for both situations and subtract. A quick units check (the answer must be a length) also kills wrong options fast.
Answer from the NTA final answer key. Solution by PrepWiser.