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The angular speed of an electron revolvi...

The angular speed of an electron revolving around the H-nucleus is proportional to

A

`1//r`

B

`1//r^(3//2)`

C

`1//r^(2)`

D

`r^(3//2)`

Text Solution

Verified by Experts

The correct Answer is:
B
We know that, therefore `1/(4pivarepsilon_(0))cdot(Ze^(2))/r^(2)=(mv^(2))/r`
therefore`v=sqrt((1/(4pivarepsilon_(0))cdot(Ze^(2))/(mr)))=k/sqrtr`
(Let `k=(Ze^(2))/(4pivarepsilon_(0)m)` )
Angular speed ,`omega=v/r=k/(sqrtrcdotr)=k/(r^(3//2))`
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