The electron in the hydrogen atom is moving with a speed at `2.3xx10^(6) m//s`in an orbit of radius 0.53 Å.The magnetic of the revolving electron is

Options

A. `9.13xx10^(-24)Am^(2)`
B. `8.7xx10^(-24)Am^(2)`
C. `9.3xx10^(-24)Am^(2)`
D. `9.3xx10^(-24) Am^(2)`

Answer

D

Explanation

Period of revolution of electron,
`T=(2pir)/(V)`
`=(2pixx0.53xx10^(-10))/(2.3xx10^(6))`
`(2.3142xx0.53xx10^(-10))/(2.3xx10^(6))`
`therefore T=1.1448xx10^(-16) s`
current,`I=(e)/(T)`
`(1.6xx10^(-19))/(1.448xx10^(-16))`
`=1.105xx10^(-3)A`
Magnetic moment of the revolving electron,
`M=IAxxIxxpir^(2)`
`therefore M=1.105xx10^(-3)xx3.142xx(0.53xx10^(-10))^(2)`
`=9.753xx10^(-24) Am^(2)`