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

Answer

D

Explanation

When electron in hydrogen atom jumps form 3rd excited state to second excited state :
`(hc)/(lambda_(1))=E_(3)-E_(4)=13.6[1/(3^(2))-1/(4^(2))]`
`=13.6xx7/144...... (i)`
When electron jumps form second excited state to first excited state, `(hc)/(lambda_(2))=E_(2)-E_(3)=13.6[1/(2^(2))-1/(3^(2))]`
`=13.6xx5/36...... (ii)`
Divide (ii) by (i), `(lambda_(1))/(lambda_(2))=(5//36)/(7//144)=20/7`