In hydrogen atom, the electron is making `6.6xx10^(15)` rev/s around the nucleus in an orbit of radius `0.528 Å`. The magnetic moment `(A-m^(2))` will be

Explanation

According to the questions ,
angular frequency `omega. =6.6xx10^(15)"revs"^(-1)`
For 1 revolution frequncy `,omega=6.63xx10^(15)xx2pi`
Radius `r=0.528Å` and change `e=1.6xx10^(-19)C`
Current with the motion of electron ,
`I=(q)/(t)=(eomega)/(2pi)=(1.6xx10^(-19)xx6.6xx10^(15)x2pi)/(2pi)" " ( :.=2pir)`
`=1.6xx10^(-19)xx6.6xx10^(156.6xx1.6xx10^(-4)`
`=1.056xx10^(-3)A`
Magnetic moment , `M =IA=1.056xx10^(-3)xxpixxr^(2)`
`=3.14xx1.056xx10^(-3)xx(0.53Å)^(2)`
`=9.314xx10^(-24)A-m^(2)~~1xx10^(-23) A^(2)`