Thank you for reporting, we will resolve it shortly
Q.
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
Atoms
Solution:
For hydrogen atom,
The kinetic energy of the electron in $n^{th}$ state is
$K = \frac{me^{4}}{8\varepsilon^{2}_{0}h^{2}n^{2}} = \frac{13.6}{n^{2}} eV$
where $\frac{me^{4}}{8\varepsilon ^{2}_{0}h^{2}} = 13.6\, eV$
The potential energy of the electron in $n^{th}$ state is
$U = - \frac{2me^{4}}{8\varepsilon ^{2}_{0}h^{2}n^{2}} = \frac{27.2}{n^{2}} eV$
Total energy of the electron in $n^{th}$ state is
$E = K +U = \frac{me^{4}}{8\varepsilon ^{2}_{0}h^{2}n^{2}}- \frac{2me^{4}}{8\varepsilon ^{2}_{0}h^{2}n^{2}}$
$= -\frac{me^{4}}{8\varepsilon ^{2}_{0}h^{2}n^{2}} = - \frac{13.6}{n^{2}} eV \quad\therefore \quad \frac{U}{E } = 2$