Calculate the angular frequency of an electron occuppying the second Bohr orbit of `He^(+)` ion.

Explanation

Velocity of electron in the nth orbit of H-like particle `(v) = (2pi Ze^(2))/(nh)`
Radius of electron in the nth orbit of H-like particle `(r_(n)) = (n^(2) h^(2))/(4pi^(2) me^(2) Z)`
Angular frequency `(omega) = (v)/(r_(n)) = (2pi Ze^(2))/(nh) xx (4pi^(2) me^(2) Z)/(n^(2) h^(2)) = (8pi^(3) Z^(2) me^(4))/(n^(3) h^(3))`
For `He^(+), Z = 2, n = 2` (Given)
Also `m = 9.108 xx 10^(-28)J, h = 6.625 xx 10^(-27)` erg sec, `e = 4.803 xx 10^(-10)` esu
Substituting these values, we get
`omega = (8 xx (22//7)^(3) xx (2)^(2) xx (9.108 xx 10^(-28)) xx (4.803 xx 10^(-10))^(4))/((2)^(3) xx (6.625 xx 10^(-27))^(3)) = 2.067 xx 10^(16) s^(-1)`