1. Use Einstein’s photoelectric equation

For a photoelectron emitted by radiation of frequency $\nu$:

$h\nu = h\nu_0 + K_{\max}$

So,

$K_{\max} = h(\nu-\nu_0)$

where $\nu_0$ is the threshold frequency.

2. Relate kinetic energy to momentum

For the emitted electron,

$K = \frac{p^2}{2m}$

Hence,

$p = \sqrt{2mK}$

Substituting $K = h(\nu-\nu_0)$:

$p = \sqrt{2m h(\nu-\nu_0)}$

3. Use de Broglie relation

The de Broglie wavelength is

$\lambda = \frac{h}{p}$

Therefore,

$\lambda = \frac{h}{\sqrt{2m h(\nu-\nu_0)}}$

Thus,

$\lambda \propto \frac{1}{\sqrt{\nu-\nu_0}}$

4. Match with the options

This corresponds to:

$\lambda \propto \frac{1}{(\nu-\nu_0)^{1/2}}$

So the correct option is D.

5. Check all options briefly

• A: $\lambda \propto \dfrac{1}{(\nu-\nu_0)^{3/2}}$ ❌
• B: $\lambda \propto \dfrac{1}{(\nu-\nu_0)^{1/4}}$ ❌
• C: $\lambda \propto \dfrac{1}{(\nu-\nu_0)}$ ❌
• D: $\lambda \propto \dfrac{1}{(\nu-\nu_0)^{1/2}}$ ✅

Hence, the derived answer matches the stored correct answer.