Matter Waves: de Broglie Wavelength question from MHT CET 2026 (12 May, Shift 2)

Answer: (b)

1. Work function $\approx 0$, so the photon energy becomes the electron's KE: $\dfrac{hc}{\lambda} = \dfrac{p^2}{2m}$.
2. With $p = h/\lambda_1$: $\dfrac{hc}{\lambda} = \dfrac{h^2}{2m\lambda_1^2}$
3. $\lambda = \dfrac{2mc}{h}\lambda_1^2 \Rightarrow \lambda \propto \lambda_1^2$
4. So the answer is (b).