The angular velocity of an electron occupying the second Bohr orbit of `He^(+)` ion is `("in" sec^(-))`:

Options

A. `2.067xx10^(16)`
B. `3.067xx10^(16)`
C. `1.067xx10^(18)`
D. `2.067xx10^(17)`

Answer

1

Explanation

Velocity of an electron in `He^(+)` ion in an orbit`=(2piZe^(2))/(nh)` …(i)
Radius of `He^(+)` ion in an orbit `=(n^(2)h^(2))/(4pi^(2)me^(2)Z)` …(ii)
By equations (i) and (ii),
Angular velocity `(omega)=u/r=(8pi^(3)Z^(2)me^(4))/(n^(3)h^(3))` ...(iii)
`=(8xx(22//7)^(3)xx(2)^(2)xx(9.108xx10^(-28))xx(4.803xx10^(-10))^(4))/((2)^(3)xx(6.626xx10^(-26))^(3))=2.067xx10^(16) sec^(-1)`