1. Use Einstein’s photoelectric equation

For the fastest emitted photoelectron, K_{max} = \frac{hc}{\lambda} - \phi_0 with the condition $\lambda < \dfrac{hc}{\phi_0}$ so that emission occurs.

2. Relate kinetic energy to de-Broglie wavelength

For the fastest electron, $p = \sqrt{2mK_{\max}}$ Hence its de-Broglie wavelength is $\lambda_d = \frac{h}{p} = \frac{h}{\sqrt{2mK_{\max}}}$ So, $\lambda_d = \frac{h}{\sqrt{2m\left(\frac{hc}{\lambda}-\phi_0\right)}}$

3. Express the kinetic energy in terms of $\lambda_d$

From $\lambda_d = \frac{h}{\sqrt{2mK_{\max}}}$ we get $K_{\max} = \frac{h^2}{2m\lambda_d^2}$

4. Differentiate to relate changes

From Einstein’s equation, $K_{\max} = \frac{hc}{\lambda} - \phi_0$ Taking differential, $dK = -\frac{hc}{\lambda^2}d\lambda$

Also, from $K = \frac{h^2}{2m\lambda_d^2}$ we get $dK = -\frac{h^2}{m\lambda_d^3}d\lambda_d$

5. Equate the two expressions for $dK$

$-\frac{h^2}{m\lambda_d^3}d\lambda_d = -\frac{hc}{\lambda^2}d\lambda$

Therefore, $\frac{d\lambda_d}{d\lambda} = \frac{mc}{h}\,\frac{\lambda_d^3}{\lambda^2}$

Thus, $\frac{\Delta \lambda_d}{\Delta \lambda} \propto \frac{\lambda_d^3}{\lambda^2}$ for small changes.

6. Match with options

This corresponds to: $\boxed{\frac{\lambda_d^3}{\lambda^2}}$ So the correct option is D.

7. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.