An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

Given:

• An electron moves perpendicular to a uniform magnetic field $B$, resulting in circular motion due to the Lorentz force.
• We assume that Bohr’s quantization is applicable.

Step 1: Magnetic Force and Circular Motion

The Lorentz force on the electron due to the magnetic field is:

$$F = q v B$$

Since the electron moves in a circular path, the centripetal force is provided by the Lorentz force:

$$\frac{m v^2}{r} = q v B$$

Rearrange to find the radius of the circular motion:

$$r = \frac{m v}{q B}$$

Step 2: Bohr’s Quantization Condition

According to Bohr’s quantization rule for angular momentum:

$$m v r = n \hbar$$

For the first excited state, we take $n = 2$:

$$m v r = 2 \hbar$$

Using $r = \frac{m v}{q B}$, substitute $v$ from the quantization condition:

$$m \left(\frac{q B r}{m} \right) r = 2 \hbar$$

 $$q B r^2 = 2 \hbar$$

Solve for $r$:

$$r = \sqrt{\frac{2 \hbar}{q B}}$$

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