Question

In the Bohr model of hydrogen atom, the ratio of the potential energy to the total energy of an electron in $n^{th}$ state is

• A. -1
• B. 1
• C. -2
• D. 2

Answer 1

Answer: (D)

For hydrogen atom,
The kinetic energy of the electron in $n^{th}$ state is
$K=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}=\frac{13.6}{n^2}eV$
where $\frac{m{e}^4}{8{\epsilon }_0^2{h}^2}=13.6eV$
The potential energy of the electron in $n^{th}$ state is
$U=-\frac{2m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}=\frac{27.2}{n^2}eV$
Total energy of the electron in $n^{th}$ state is
$E=K+U=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}-\frac{2m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$
$=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}=-\frac{13.6}{n^2}eV\therefore \frac{U}{E}=2$