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

Options

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

Answer

D

Explanation

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