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The electron in the hydrogen atom ...

The electron in the hydrogen atom is moving with a speed of `2.5 xx10^(6)` m/s in an orbit of radius 0.5 A what is the magnetic moment of the revolving electron ?

A

`4.64xx10^(-226) Am^(2)`

B

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

C

`4.64 xx10^(-245) Am^(2)`

D

`9.28 xx10^(-22) Am^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C

The magneitc moment of the revolving electron is given by `M_(0) =("neh")/(4pim_(e ))=("neh")/(4pim_(e ))`
Thus `M_(0) prop n` (the principla quantum number)
The magnetic moment of the revolving electron is
`M=IA =(e )/(T)xxpir^(2)` but `T=(2pir)/(v) therefore M=(ev)/(2pir)xxpir^(2)=("evr")/(2)`
`=1.6xx2.5xx0.25xx10^(-23)`
`=10^(-23)Am^(2)`
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