JEE Physics 2025 Question with Solution

Given: Incident radiation has wavelength $\lambda_i$, threshold wavelength is $\lambda_0$, and the emitted electron has de-Broglie wavelength $\lambda_e$.

Find: The correct expression for $\lambda_e$.

Use Einstein's photoelectric equation and the de-Broglie relation.

The energy of the incident photon is

$$E_{\text{incident}} = \frac{hc}{\lambda_i}$$

and the work function is

$$\phi = \frac{hc}{\lambda_0}$$

So the maximum kinetic energy of the emitted electron is

$$K_{\max} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}$$

Now use

$$K_{\max} = \frac{p^2}{2m}$$

Therefore,

$$p = \sqrt{2mK_{\max}} = \sqrt{2m \left( \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} \right)}$$

The de-Broglie wavelength is

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

Hence,

$$\lambda_e = \frac{h}{\sqrt{2m \left( \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} \right)}}$$

which can be written as

$$\lambda_e = \frac{h}{\sqrt{2mc \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_0} \right)}}$$

Therefore, the correct option is A.

Given: $K_{\max} = \frac{hc}{\lambda_i} - \phi$ and $\phi = \frac{hc}{\lambda_0}$.

Find: The expression for electron de-Broglie wavelength.

First substitute the threshold condition into the photoelectric equation:

$$K_{\max} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} = hc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)$$

Now relate kinetic energy and momentum:

$$K_{\max} = \frac{p^2}{2m}$$

So,

$$p = \sqrt{2mK_{\max}}$$

Substitute for $K_{\max}$:

$$p = \sqrt{2mch\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}$$

Then the de-Broglie wavelength becomes

$$\lambda_e = \frac{h}{\sqrt{2mch\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}}$$

This is algebraically the same as option A because

$$h\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right) = \frac{h}{\lambda_i} - \frac{h}{\lambda_0}$$

So option C shows the same derived form, but the solution identifies A as the correct option and option A is the equivalent expression presented in the list.

Therefore, the correct option is A.