JEE Main 2021 · 27 August · evening · Single correct · +4 / −1
A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?
- (a) 0.96 V
- (b) 1.25 V
- (c) 0.24 V
- (d) 1.5 V
Answer and solution
Answer (b) 1.25 V
Solution
Physics sir · 4 steps
Write Einstein's equation for both lamps and subtract: the work function drops out.
- Einstein's equation. For each source, eV₀ = hc/λ - φ, where φ is the work function of the same material. Subtract the two equations and φ cancels.
e(V₂ - V₁) = hc(1/λ₂ - 1/λ₁)
- Photon energies. Use hc = 1240 eV nm. For 474.6 nm, E₂ = 1240/474.6 = 2.613 eV. For 670.5 nm, E₁ = 1240/670.5 = 1.849 eV. The difference is 0.763 eV.
- New stopping voltage. V₂ = 0.48 + 0.763 = 1.24 V. As a check, φ = 1.849 - 0.48 = 1.37 eV, and 2.613 - 1.37 = 1.24 eV gives the same result.
- So. The stopping voltage rises to about 1.24 V, closest to 1.25 V, option (b).
The idea to keep. Same metal, two wavelengths? Subtract the two Einstein equations. The change in stopping voltage equals the change in photon energy in eV, and hc = 1240 eV nm makes that quick.
Answer from the NTA final answer key. Solution by PrepWiser.