Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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A Uniform Magnetic Field B Exist in a Region. an Electron Projected Perpendicular to the Field Goes in a Circle. Assuming Bohr'S Quantization - Physics

Sum

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.

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Solution

According to Bohr's quantization rule,
`mvr = (nh)/(2pi)`

'r' is minimum when 'n' has minimum value, i.e. 1.

`mv = (nh)/(2pir)` ........(1)

Again,` r = (mv)/(qB)`

`rArr mv = rqB`      .........(2)

From (1) and (2), we get

`rqB = (nh)/(2pir) `        [From (1)]

`rArr r^2 = (nh)/(2pieB)`  [∴q=e ]

`rArr r = sqrt((h)/(2pieB)`  [n = 1]

(b) For the radius of nth orbit,

`r = sqrt((nh)/(2pieB)`

`(C)  mvr = (nh)/(2pi), r = (mv)/(qB)`

Substituting the value of 'r' in (1), we get

`mvxx(mv)/(qB)=(nh)/(2pi)`

`rArr m ^2v^2 = (heB)/(2pi)`     [n=1,q=e]

`v^2 = (heB)/(2pim^2)`

`v = sqrt((heB)/(2pim^2)`

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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 21 Bohr’s Model and Physics of Atom
Q 44 | Page 386
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