A uniform magnetic field B exists in a region. An electrons 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 electrons (b) the radius of the nth orbit and (c) the minimum possible speed of the electron.

Answer

A, B, C

Explanation

According to Bohr's quantization rule
`mvr = (nh)/(2pi)`
'r' is less when 'n' has least value i.e. 1
or `mv=(nh)/(2pir)`…(1)
Again, `r=(mv)/(qB)` …(2)
From (1) and (2) `rqB=(nh)/(2pir)` [From (1)]
implies `r^2 = (nh)/(2pireB)` [:'`9=e`]
implies `r=sqrt((h)/(2pi)eB)`[ Hence `n=1`]
(b) For the radius of `n^(th)` orbit,
`r=sqrt ((nh)/(2pieB))`
(c) `mvr = (nh)/(2pi)` `r=(mv)/(qB)`
Substituting the value of 'r' in (1)
`mv xx (mv)/(qB) = (nh)/(2pi)`
implies `m^2v^2 = (nheB)/(2pi)` [`n=1 q=e`]
implies `v^2 = (heB)/(2pim^2)`
or `v=(heB)/(2pim^2)`