Question

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

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

Answer 1

Answer: (C)

0.967 eV

Energy of $nth$ state of Hydrogen atom, $E_n=-\frac{13.6}{n^2}$ $eV$

Energy radiated during the transition, $E=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ $eV$

$4th$ excited state means $n_2=5$ and $2nd$ excited state means $n_1=3$

$E=13.6(\frac{1}{3^2}-\frac{1}{5^2})$ $eV$

$⟹E=0.967$ $eV$