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{me^{2}}{r}$
• D. $ -\frac{1}{2} \frac{e^{2}}{r}$

Answer 1

Answer: (D)

$= PE + KE \,\,\,\,\left\{\because \frac{ mv ^{2}}{ r }=\frac{ ke ^{2}}{ r ^{2}}\right.$

$\left.=-\frac{ ke ^{2}}{ r }+\frac{1}{2} mv ^{2} \,\,\,\,\therefore \frac{1}{2} mv ^{2}=\frac{1}{2} \frac{ ke ^{2}}{ r }\right\}$

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

$=-\frac{1}{2} \frac{ ke ^{2}}{ r }$

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