Question

In the Bohr's model of hydrogen atom, the ratio of the kinetic energy to the total energy of the $\rho$ in the $nth$ quantum state is:

• A. 2
• B. -2
• C. +1
• D. -1

Answer 1

Answer: (D)

The energy $E$ of an electron in an orbit is the sum of kinetic and potential energies. The kinetic energy of the electron is
$K=\frac{1}{2}m{v}^2=\frac{Z{e}^2}{8\pi {\epsilon }_{o}r}$
where $Z$ is atomic number, $e$ is charge, $r$ is radius.
The potential energy of an electron in an orbit of radius $r$ due to electrostatic attraction by the nucleus is given by
$U=\frac{1}{4\pi {\epsilon }_0}\frac{(Ze)(-e)}{r}$
$=-\frac{1}{4\pi {\epsilon }_o}\frac{Z{e}^2}{r}$
Total energy of the electron is therefore,
$E=K+U=\frac{Z{e}^2}{8\pi {\epsilon }_{o}r}-\frac{Z{e}^2}{4\pi {\epsilon }_{0}r}=-\frac{Z{e}^2}{8\pi {\epsilon }_{0}r}$
$\therefore \frac{K}{E}=-1$