Question

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

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

Answer 1

Answer: (A)

$\frac{4 \pi^{2}mr^{2}}{n h}$

Answer 2

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