An electron jumps from the fourth orbit to the second orbit hydrogen atom. Given the Rydberg's constant `R = 10^(7) cm^(-1)`. The frequecny , in `Hz` , of the emitted radiation will be

Options

A. `(3)/(16) xx 10^(5)`
B. `(3)/(6) xx 10^(15)`
C. `(9)/(16) xx 10^(5)`
D. `(9)/(16) xx 10^(15)`

Answer

D

Explanation

`(1)/(lambda) = R [(1)/(2^(2)) - (1)/(4^(2))]`
or `(f)/( c) = R [(1)/(4) - (1)/(16)]`
or `f = c R [(1)/(4) - (1)/(16)]`
`= 3 xx 10^(8) xx 10^(7) xx (3)/(16)`
`= (9)/(16) xx 10^(15) Hz`