The electron in hydrogen atom moves with a speed of `2.2xx10^(6)m//s` in an orbit of radius `5.3xx10^(-11)cm`. Find the magnetic moment of the orbiting electron.

Options

A. `8.3 xx 10^(-23)A-m^(2)`
B. `9.3 xx 10^(-24)A-m^(2)`
C. `7.2 xx 10^(-24)A-m^(2)`
D. `6 xx 10^(-24)A-m^(2)`

Answer

B

Explanation

Frequency of revolution , ` f=(v)/( 2pir )`
The moving charge is equivalent to a current loop , given by
`i=f xx e or i=(ev)/( 2pir )`
If A be the area of the orbit , then the magnetic moment of the orbiting electron is ,
`M=iA=((ev)/( 2pi r))(pir^(2))=(evr)/(2)`
Putting the values , we have
`M=((1.6 xx 10^(-19))(2.2 xx 10^(6))(5.3 xx 10^(-11)))/(2)`
`=9.3 xx 10^(-24)A-m^(2)`