Some energy levels of a molecule are shown in the fig. The ratio of the wavelengths `r=lambda_(1)//lambda_(2)`, is given by

Options

A. `r=3/4`
B. `r=1/3`
C. `r=4/3`
D. `r=2/3`

Answer

B

Explanation

As is clear form fig.
`(hc)/(lambda_(1))=-E-(-2E)=E.......(i)`
and `(hc)/(lambda_(2))=-E-(-4/3 E)=E/3.......(ii)`

Dividing (ii) by (i), we get
`(lambda_(1))/(lambda_(2))=(E//3)/E=1/3`