Excited atoms emit radiations consisting of only certain discrete frequencies or wavelengths. In spectroscopy it is often more convenient to use frequencies or wave numbers than wavelength because frequencies and wave numbers are proportional to energy and spectroscopy involves transitions between different energy levels. The line spectrum shown by a mono electronic excited atom (a finger print of an atom) is called atomic spectrum. `1/(lambda) = Z^2 R [1/(n_1^2 - 1/(n_2^2)]`
The given diagram indicates the energy levels of certain atom. When an electron moves from 2E level to E level, a photon of wavelength `lambda` is emitted. The wavelength of photon emitted during its transition from `(4E)/(3)`level to E level is

Options

A. `lambda/3`
B. `(3 lambda)/(4)`
C. `(4lambda)/(3)`
D. `3lambda`

Answer

D

Explanation

hv = 2E -E
`(hc)/(lambda) = E implies lambda = (hc)/(E) ` Also, `(hc)/(lambda_1) = (4E)/(3) - E = E/3`
`E= 3 lambda`