An electron jumps from the `4th` orbit to the `2nd` orbit of hydrogen atom. Given the Rydberg's constant `R = 10^(5) cm^(-1)`. The frequency in `Hz` of the emitted radiation will be

Options

A. `(3)/(16) xx 10^(5)`
B. `(3)/(16) xx 10^(15)`
C. `(9)/(16) xx 10^(15)`
D. `(3)/(4) xx 10^(15)`

Answer

C

Explanation

`n_(1)= 2, n_(2)=4`
`f= Rc [(1)/(n_(1)^(2))- (1)/(n_(2)^(2))]`
`rArr f= 10^(7) xx 3 xx 10^(8) [(1)/(2^(2))- (1)/(4^(2))]`
`rArr f= (3 xx 10^(15) xx 3)/(16) rArr f= (9)/(16) xx 10^(15)Hz`