Q33 / 155 · WBJEE 2016 · Chemistry

The time taken for an electron to complete one revolution in Bohr orbit of hydrogen atom is

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

Solution

Let the distance travelled in $T$ time $=2 \pi r$ (Circumference of orbit) $\therefore \quad$ Velocity $(v)=\frac{2 \pi r}{T}$ Also, $v=\frac{n \cdot h}{2 \pi m r}$ $\therefore \quad \frac{1}{T \text { (Time period) }}=$ frequency $v=\text{v} \times \frac{1}{2 \pi r}$ From Eq. $$ \begin{array}{l} \mathrm{v}=\frac{1}{T}=\frac{n h}{2 \pi \times 2 \pi m r} \\ v=\frac{n h}{4 \pi^{2} r^{2} m} \end{array} $$ $T$ (time taken in one revolution) $$ =\frac{4 \pi^{2} r^{2} m}{n h} $$

Hence, the correct answer is option: 'C'.

Relevant Questions