An atom makes a transition from a state of energy `E` to one of lower energy `E`. Which of the following gives the wavelength of the radiation emitted in terms of the Planck's constants `h` and the speed of light `c`?

Options

A. `(E_(2) - E_(1))/(hc)`
B. `(hc)/(E_(2)) - (hc)/(E_(1))`
C. `(hc)/(E_(1))- (hc)/(E_(2))`
D. `(hc)/(E_(2) - E_(1))`

Answer

D

Explanation

By quantum theory of radiation, the energy change `E` between enegry levels is proportional to the frequency of electromagnetic radiation `f` and is given by the quantion below
`E = hf = (hc)/(lambda)`
where `h` is the Planck's constant, `c` is the speed of light and `lambda` is the wavelength of the radiation. ltbRgt Now, energy change by transition from `E_(3)` to `E_(1)` is given by
`E = E_(2) = E_(1) ....(1)`
From `(1) and (2)`, we have
`E_(2)-E_(1) = (hc)/(lambda)`
`rArr lambda = (hc)/((E_(2)-E_(1)))`