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. `2lambda`
D. `3lambda`

Answer

4

Explanation

For II to I transition, `DeltaE=(4E)/3-E=(hc)/(.^(lambda)II rarr I), E/3=(hc)/(.^(lambda)II-I)`
For III to I transition, `DeltaE=2E-E=(hc)/lambda` or `E=(hc)/lambda :. (hc)/(3xxlambda)=(hc)/(.^(lambda)II-I) lambda_(II-I)=3lambda`