Question

A uniform magnetic field B exist in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the minimum possible speed of the electron.

Answer 1

According to Bohr's quantization rule,
$mvr=\frac{nh}{2\mathrm{\pi }}$
'r' is minimum when 'n' has minimum value, i.e. 1.
$$\begin{gathered} mv=\frac{nh}{2\mathrm{\pi }r}...\left(1\right) \\ \mathrm{Again},r=\frac{mv}{qB} \\ \Rightarrow mv=rq\mathrm{B}...\left(2\right) \end{gathered}$$

From (1) and (2), we get
$$\begin{gathered} rq\mathrm{B}=\frac{nh}{2\mathrm{\pi }r}\left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow r^2=\frac{nh}{2\mathrm{\pi }e\mathrm{B}}\left[\therefore q=e\right] \\ \Rightarrow r=\sqrt{\frac{h}{2\mathrm{\pi eB}}}\left[n=1\right] \end{gathered}$$

(b) For the radius of n^th orbit,
$r=\sqrt{\frac{nh}{2\mathrm{\pi }e\mathrm{B}}}$

(c) $mvr=\frac{nh}{2\mathrm{\pi }},r=\frac{mv}{q\mathrm{B}}$
Substituting the value of 'r' in (1), we get
$$\begin{gathered} mv\times \frac{mv}{q\mathrm{B}}=\frac{nh}{2\mathrm{\pi }} \\ \Rightarrow m^2{v}^2=\frac{he\mathrm{B}}{2\mathrm{\pi }}\left[n=1,q=e\right] \\ \Rightarrow v^2=\frac{he\mathrm{B}}{2\mathrm{\pi }{m}^2} \\ v=\sqrt{\frac{he\mathrm{B}}{2\mathrm{\pi }{m}^2}} \end{gathered}$$