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

Answer 1

Answer: (D)

$=PE+KE\{\because \frac{m{v}^2}{r}=\frac{k{e}^2}{r^2}$

$=-\frac{k{e}^2}{r}+\frac{1}{2}m{v}^2\therefore \frac{1}{2}m{v}^2=\frac{1}{2}\frac{k{e}^2}{r}\}$

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

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

$=-\frac{1e^2}{2r}[$ In CGS system $K=1]$