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)

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

Answer 2

= 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 $K =1]$