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. `8.27 xx 10^(-26) Am^(2)`
B. `9.27 xx 10^(-27) Am^(2)`
C. `9.3 xx 10^(-26)Am^(2)`
D. `8.8 xx 10^(-27)Am^(2)`

Answer

C

Explanation

Frequency of revolution , ` f=(v)/( 2pir )`
The moving charge is equivalent to a current loop , given by
`l=f xx e or l=(ev)/( 2pir )`
If A be the area of the orbit , then the magnetic moment of the orbiting electron is ,
`M=lA=((ev)/( 2pi r))(pir^(2))=(evr)/(2)`
Putting the values , we get
`M=((1.6 xx 10^(-19))(2.2 xx 10^(6))(5.3 xx 10^(-11) xx 10^(-2)))/(2)`
`=9.3 xx 10^(-26)A-m^(2)`