Question

Time taken by an electron to complete one revolution in the Bohr orbit of hydrogen atom in terms of its radius r and quantum number n is:

• A. $\frac{4{\pi }^{2}m{r}^2}{nh}$
• B. $\frac{2\pi r}{nh}$
• C. $\frac{{\pi }^2{r}^2}{8\pi r^2}$
• D. $\frac{n{h}^2}{8\pi r}$

Answer 1

The correct option is A $\frac{4{\pi }^{2}m{r}^2}{nh}$

We need to find time taken by an $e^{-}$ to complete one revolution in Bohr orbit of $H$-atom in terms of radius ${}^{\prime }{r}^{\prime }$ and quantum no. ${}^{\prime }{n}^{\prime }$.
We have two relations; $T_n=\frac{2\pi r_n}{{\text{v}}_n}...\left(i\right)$
$m{\text{v}}_n{r}_n=\frac{nh}{2\pi }...\left(ii\right)$
where, $T_n=\text{Time period}$
$r_n=\text{radius}$
${\text{v}}_n=\text{velocity}$
$n=\text{Orbit number}$
From equation (ii),
${\text{v}}_n=\frac{nh}{2\pi m{r}_n}...\left(iii\right)$
Putting this value in equation (i),
$T_n=\frac{2\pi r_n}{\frac{nh}{2\pi m{r}_n}}=\frac{4{\pi }^{2}m{r}^2}{nh}$