Question

An electron jumps from `5th` orbit to `4th` orbit of hydrogen atom. Taking the Rydberg constant as `10^(7)` per meter. What will be the frequency of radiation emitted ?
A. `6.75 xx 10^(12) Hz`
B. `6.75 xx 10^(14) Hz`
C. `6.75 xx 10^(13) Hz`
D. None of these

Answer 1

Correct Answer - C
By using `v = RC [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
`rArr v = 10^(7) xx (3 xx 10^(8)) [(1)/(4^(2)) - (1)/(5^(2))] = 6.75 xx 10^(13) Hz`