Question

the ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is:-

• A. $1:1$
• B. $1:-1$
• C. $2:-1$
• D. $1:-2$

Answer 1

The correct option is B $1:-1$

According to Bohr’s model, the electrostatic force is balanced by the centripetal force as

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

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

$so,KE=\frac{1}{2}\frac{kZ{e}^2}{r}..........\left(1\right)$

$P.E=F.dr=-\frac{kZ{e}^2}{r}........\left(2\right)$

$\therefore totalenergybecomes$

$TE=KE+PE$

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

$TE=-\frac{1}{2}\frac{kZ{e}^2}{r}................\left(3\right)$

$\therefore \frac{KE}{TE}=\frac{1}{-1}$

$ratiois1:-1$