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. `9/16xx10^15 Hz`
B. `3/16xx10^5 Hz`
C. `3/14xx10^15`
D. `3/4xx10^15 Hz`

Answer

A

Explanation

`1/lamda = R (1/2^2 - 1/4^2) = R(1/4-1/16) = (3R)/(16)`
`therefore lamda = (16)/(3R) = 16/3 xx10^(-5) cm = 16/3 xx10^(-7) m`
`therefore v=C/lamda = (3xx10^8)/(16/3xx10^(-7)) = 9/16xx10^15 Hz`