For the hydrogen atom the energy of radiation emitted in the transitation from 4th excited state
to 2nd exicited state according to Bohr 's theory is

Options

A. 0.567 eV
B. 0.667 eV
C. 0.967 eV
D. 1.267 eV

Answer

C

Explanation

For the hydrogen atom , for the ground state n =1 , for the second excited state n=3 and for the 4th excited state n=5
and `E_n = - (13.6)/(n^2), E_3 = - (13.6)/(9) " and " E_5 = - (13.6)/(25)`
`therefore E_5-E_3 = (13.6)/(9) - (13.6)/(25) = 13.6 [1/9 - 1/25]`
`= (13.6xx16)/(225) = 0.967 eV`
This is the energy of the emitted radiation in the given transition .