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 :

When an electron is projected perpendicular to a uniform magnetic field $$B$$, it moves in a circular path due to the magnetic force acting as the centripetal force. The force balance equation is given by:

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

where $$e$$ is the charge of the electron, $$v$$ is its speed, $$m$$ is its mass, and $$r$$ is the radius of the circular path. Simplifying this equation:

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

Rearranging, we get:

$$ m v = e B r \quad \text{(1)} $$

Bohr's quantization condition states that the angular momentum is quantized and given by:

$$ m v r = n \frac{h}{2\pi} \quad \text{(2)} $$

where $$n$$ is the principal quantum number and $$h$$ is Planck's constant. The first excited state corresponds to $$n = 2$$ (since the ground state is $$n = 1$$). Substituting $$n = 2$$ into equation (2):

$$ m v r = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} \quad \text{(3)} $$

We now have two equations: equation (1) $$m v = e B r$$ and equation (3) $$m v r = \frac{h}{\pi}$$. Substituting equation (1) into equation (3):

$$ (e B r) \times r = \frac{h}{\pi} $$

Simplifying:

$$ e B r^2 = \frac{h}{\pi} $$

Solving for $$r^2$$:

$$ r^2 = \frac{h}{\pi e B} $$

Taking the square root of both sides:

$$ r = \sqrt{\frac{h}{\pi e B}} $$

This is the radius of the electronic orbit in the first excited state. Comparing with the given options, this expression matches option A, $$\sqrt{\frac{h}{\pi eeB}}$$, when interpreted as $$\sqrt{\frac{h}{\pi e B}}$$ (considering "ee" as a typographical error for "e").

Thus, the correct option is A.