A hydrogen-like atom emits rediationof frequency `2.7 xx 10^(15)` Hz` when if makes a transition from `n = 2 to n = 1`. The frequency emitted in a transition from `n = 3 to n = 1` will be

Options

A. `1.8 xx 10^(15) Hz`
B. `3.2 xx 10^(15) Hz`
C. `4.7 xx 10^(5) Hz`
D. `6.9 xx 10^(15) Hz`

Answer

B

Explanation

`f = c Z^(2) R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
` implies 2.7 xx 10^(15) = c z ^(2) R [(1)/(1^(2)) - (1)/(2^(2))]`
`f' = c z ^(2) R [(1)/(1^(2)) - (1)/(3^(2))]`
Divide and solve to get `f = 3.2 xx 10^(15) H z`