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. `3.2xx10^(15)Hz`
B. `32xx10^(15)Hz`
C. `1.6xx10^(15)Hz`
D. `16xx10^(15)Hz`

Answer

A

Explanation

The frequency v of the emitted electromagnetic radiation, when a hydrogen atom de-excites from level `n_(2)` to `n_(1)` is
`v=RcZ^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
When transition takes place from `n_(2)=2` to `n_(1)`, then
`2.7xx10^(15)=RcZ^(2)((1)/(1^(2))-(1)/(2^(2)))`
When transition takes places from `n _(2)=3` to `n_(1)=1`, let frequency be v.
`therefore v=RcZ^(2)((1)/(1^(2)-(1)/(3^(2))))` . . .(ii)
`v=(32xx2.7xx10^(15))/(27)=3.2xx10^(15)Hz`