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

Explanation:

As per the quantization rule of Bohr,

mvr = nh2π, when ‘n’ has minimum value, then ‘r’ is minimum, that is, 1.

mv = nh2πr                       …1

Again, r = mvqB

⇒ mv = rqB                         …2

From (1) and (2), we obtain

rqB = nh2πr                          

From 1 ⇒ r2 = nh2πeB  

Therefore, q = e

⇒r = h2πeB  

n=1

(b) For the nth orbit radius,

r = nh2πeB

(c)

mvr = nh2π, r = mvqB

When the value of r is substituted in (1), we obtain

mv × mvqB = nh2π

⇒m2v2 = heB2π                    

n=1, q=e

v = heB2πm2