Chapter 27, Problem 62P

To determine

To show:

The magnitude of the electrostatic potential energy of an electron in a hydrogen atom is twice the magnitude of its kinetic energy.

Answer to Problem 62P

Solution:

The magnitude of the electrostatic potential energy of an electron in a hydrogen atom is twice the magnitude of its kinetic energy.

Explanation of Solution

Calculation:

The expression for the potential energy of an electron in a hydrogen atom can be given as:

$\text{PE}=-\frac{kZ{e}^2}{r}$_____________________________________________(1)

Where, k is Coulombs constant

Z is the atomic number, that is 1

e is the charge of electron

and, r is the radius of Bohr orbit

the electrostatic force of attraction between the nucleus and the revolving electron is equal the the centrifugal force, the relation can be written as:

$\frac{m{v}^2}{r}=-\frac{kZ{e}^2}{r^2}$

$m{v}^2=-\frac{kZ{e}^2}{r}$

Or we can say that

$\begin{array}{l}\frac{1}{2}m{v}^2=-\frac{kZ{e}^2}{2r}\\ \text{KE}=\frac{1}{2}\text{PE}\\ \text{or,}\text{PE}\text{=}\text{2KE}\end{array}$____________________________________________by eqn (1)