JEE Physics 2025 Question with Solution

Given: A monochromatic light falls on a metallic plate of work function $\phi$. The emitted electron has maximum kinetic energy and enters a magnetic field perpendicular to its velocity.

Find: The distance $AB$ between the emission point and the point where the electron hits the plate again.

From Einstein's photoelectric equation,

$$K_{\max} = \frac{hc}{\lambda} - \phi$$

The momentum of the electron is

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

In a magnetic field, the electron follows a circular path. Since it returns to the plate, it travels a semicircle, so the distance $AB$ is the diameter:

$$AB = 2R$$

Using the relation for circular motion in a magnetic field,

$$R = \frac{p}{eB}$$

Therefore,

$$AB = 2 \cdot \frac{p}{eB} = \frac{2}{eB}\sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)}$$

Hence,

$$AB = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB}$$

Therefore, the correct option is C.

Given: The maximum kinetic energy of the photoelectron is determined by the photoelectric effect, and the magnetic field is perpendicular to the initial velocity.

Find: The distance between points A and B.

1. By Einstein's photoelectric equation,

$$K_{\max} = h\nu - \phi = \frac{hc}{\lambda} - \phi$$

2. Using kinetic energy,

$$K_{\max} = \frac{1}{2}mv^2$$

So,

$$v = \sqrt{\frac{2K_{\max}}{m}}$$

Substituting for $K_{\max}$,

$$v = \sqrt{\frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right)}$$

3. In the magnetic field, the magnetic force provides centripetal force. Therefore, the radius of the circular path is

$$r = \frac{mv}{eB}$$

Substitute the value of $v$:

$$r = \frac{m}{eB} \sqrt{\frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right)}$$

Simplifying,

$$r = \frac{1}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)}$$

4. Since the electron returns to the plate, the trajectory is a semicircle. Hence,

$$AB = 2r$$

So,

$$AB = 2 \times \frac{1}{eB} \sqrt{2m \left( \frac{hc}{\lambda} - \phi \right)}$$ $$AB = \frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB}$$

Therefore, the distance between A and B is $\frac{\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right)}}{eB}$, so the correct option is C.