JEE Physics 2025 Question with Solution

Given: An electron moves perpendicular to a uniform magnetic field $B$ in a circular path, and Bohr quantization applies.

Find: The radius of the electronic orbit in the first excited state.

From magnetic force providing centripetal force,

$$\frac{mv^2}{r} = evB$$

So,

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

Using Bohr's quantization condition,

$$mvr = n\frac{h}{2\pi}$$

Substitute $mv = eBr$ from the force relation into the quantization condition:

$$eBr^2 = n\frac{h}{2\pi}$$

Hence,

$$r^2 = \frac{nh}{2\pi eB}$$

Therefore,

$$r = \sqrt{\frac{nh}{2\pi eB}}$$

For the first excited state, $n = 2$. Thus,

$$r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}}$$

This matches option C from the algebra shown in the working, although the solution marks option D as correct and concludes $\sqrt{\frac{h}{2\pi eB}}$. Following the solution's stated correct option, the marked answer is D.

Given: Circular motion of an electron in a magnetic field $B$ with Bohr quantization.

Find: Radius in the first excited state.

From the circular motion condition,

$$\frac{mv^2}{r} = evB$$

which gives

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

Now apply Bohr quantization,

$$mvr = n\hbar = \frac{nh}{2\pi}$$

Using

$$v = \frac{nh}{2\pi mr}$$

substitute into

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

Then,

$$r = \frac{m}{eB} \times \frac{nh}{2\pi mr}$$

So,

$$r^2 = \frac{nh}{2\pi eB}$$

and hence,

$$r = \sqrt{\frac{nh}{2\pi eB}}$$

For the first excited state,

$$n = 2$$

therefore,

$$r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}}$$

So the derivation leads to option C, while the solution's explicitly states option D. The extracted answer is kept as D because the solution is treated as the authority for the marked answer.