Question

If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

• A. $\frac{1}{2}\frac{k{e}^2}{r}$
• B. $-\frac{k{e}^2}{r}$
• C. $\frac{mk{e}^2}{r}$
• D. $-\frac{1}{2}\frac{k{e}^2}{r}$

Answer 1

The correct option is D $-\frac{1}{2}\frac{k{e}^2}{r}$

Total energy $=K.E.+P.E.$

K.E. $=-\frac{1}{2}m{v}^2$ = $\frac{1}{2}\frac{Zk{e}^2}{r}$ (cenrtipetal force)

P.E. $=-\frac{Zk{e}^2}{r}$

T.E. $=\frac{1}{2}\frac{Zk{e}^2}{r}$ + $-\frac{Zk{e}^2}{r}$

$=-\frac{1}{2}\frac{Zk{e}^2}{r}$

$=-\frac{1}{2}\frac{k{e}^2}{r}$ ....... ($Z=1$ for hydrogen)

Hence, option D is correct.