Question

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

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

Answer 1

The correct option is C $0.967eV$

$4^{th}$ excited state means $n=5$ and $2^{nd}$ excited state means $n=3$.

Energy of the radiation emitted due to transition from $n_2=5$ to $n_1=3$ is $E=13.6[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$ $eV$

$\therefore$ $E=13.6[\frac{1}{3^2}-\frac{1}{5^2}]$

$⟹$ $E=13.6\times \frac{16}{9\times 25}=0.967eV$