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)xx10^(5)`
B. `(3)/(16)xx10^(15)`
C. `(9)/(16)xx10^(15)`
D. `(3)/(4)xx10^(15)`

Answer

C

Explanation

As, `v=(c)/(lamda)=c.R((1)/(n_(1)^(2)-(1)/(n_(2)^(2))))=3xx10^(8)xx10^(7)((1)/(2^(2))-(1)/(4^(2)))`
`=(9)/(16)xx10^(15)`Hz