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The time taken for an electron to complete one revolution in Bohr orbit of hydrogen atom is
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2456 Upvotes
Verified Answer
The correct answer is:
$\frac{4 \pi^{2} m r^{2}}{n h}$
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)
$$
=\frac{4 \pi^{2} r^{2} m}{n h}
$$
$\therefore \quad$ 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)
$$
=\frac{4 \pi^{2} r^{2} m}{n h}
$$
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