Calculate the angular frequency of an electron occapying the second Bohr orbit of `He^(Theta)` ion

Explanation

Velocity of electron in `He^(o+)` ion in an orbit
`n = (2pi Ze^(2))/(nh)`
Radius of `He^(o+)` ion in an orbit
`r_(n) = (n^(2)h^(2))/(4pi^(2)me^(2)Z)`
Angular frequancy or angular velocity
`omega = (u)/(r_(n)) = (2pi ZE^(2) xx 4pi^(2) me^(2)Z)/(nh xx n^(2)h^(2)) = (8 pi^(2)Z^(2)me^(4))/(n^(3)h^(3))`
`:' n = 2, m = 9.108 xx 10^(-28)g ,Z = 2, h= 6.625 xx 10^(-27)`
`= 2.067 xx 10^(16)s^(-1)`
`therefore omega=(8xx(22//7)^(3)xx(2)^(2)xx9.108xx10^(-28)xx(4.803xx10^(-10))^(4))/((2)^(3)xx(6.625xx10^(-27))^(3))`
`=2.067xx10^(16)s^(-1)`