The energy of a I,II and III energy levels of a certain atom are `E,(4E)/(3)` and 2E respectively. A photon of wavelength `lambda` is emitted during a transition from III to I. what will be the wavelength of emission for II to I?

Options

A. `(lambda)/(2)`
B. `lambda`
C. `2 lambda`
D. `3 lambda`

Answer

D

Explanation

For II to I transition
`DeltaE =(4E)/(3) - E = (hc)/(lambda_(II to I)) , (E)/(3) = (hc)/(lambda_(II -I))`
For III to I transition .
`Delta E = 2E = (hc)/(lambda) or E = (hc)/(lambda)`
`therefore (hc)/(3 xx lambda) = (hc)/(lambda_(II -I)) = lambda_(II - I) = 3 lambda`