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An electron is revolving in an orbit of ...

An electron is revolving in an orbit of radius `5.0xx10^(-11)` m with a speed of `2.0xx10^4m//s`. The magnetic moment associated with the atom will be

A

`12xx10^(-26)Am^2`

B

`8xx10^(-26)Am^2`

C

`9xx10^(-26)Am^2`

D

`6xx10^(-26)Am^2`

Text Solution

Verified by Experts

The correct Answer is:
B
`M=ixxA=(f.e)pir^2=omega/(2pi).e.pir^2," ""ver"/2=M`
`thereforeM=(1.6xx10^(-19)xx2xx10^4xx5xx10^(-11))/2=8xx10^(-2)Am^2" "[T=(2pir)/v]`
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