Question

Time taken for an electron to complete one revolution in the Bohr orbit of hydrogen atom is

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

Answer 1

Answer: (A)

By Bohr postulate, $m\upsilon r=n\frac{h}{2\pi }or\upsilon =\frac{nh}{2\pi mr}$
No. of revolutions per sec
$=\frac{Velocity}{Circumferenceoftheorbit}=\frac{\upsilon }{2\pi r}$
Substituting value of $\upsilon ,$ we get
$=\frac{nh}{2\pi mr}\times \frac{1}{2\pi r}=\frac{nh}{4{\pi }^{2}m{r}^2}$
$\frac{nh}{4{\pi }^{2}m{r}^2}$ revolutions completed in 1s
$\therefore$ 1 revolution will take $\frac{1}{\frac{nh}{4{\pi }^{2}m{r}^2}}=\frac{4{\pi }^{2}m{r}^2}{nh}$