Electron in hydrogen atom first jumps from third excited state to second excited state and then form second excited state to first excited state. The ratio of wavelength `lambda_(1): lambda_(2)` emitted in two cases is

Options

A. `27//5`
B. `20//7`
C. `7//5`
D. `27//20`

Answer

B

Explanation

Wavelength observed from transition of `n_(i) " to " n_(f) " is " lamda=(1)/([(1)/(n_(f)^(2))-(1)/(n_(i)^(2))])`
For `lamda_(1), n_(i)=4, n_(f)=3`
For `lamda_(2),n_(i)=3,n_(f)=2`
We get
`lamda_(1) : lamda_(2)=20 : 7`