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An electron is moving in Bohr's fourth orbit. Its de-Broglie wave length is $\lambda$. What is the circumference of the fourth orbit?
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Verified Answer
The correct answer is:
$4 \lambda$
According to Bohr's concept, an electron always move in the orbit with angular momentum ( $m v r)$ equal to $n h / 2 \pi$.
$\therefore \quad m v r=\frac{n h}{2 \pi}$
or
$r=\frac{n}{2 \pi} \cdot\left(\frac{h}{m v}\right) \text { or } r=\frac{n \lambda}{2 \pi}$
(From de -Broglie equation, $\lambda=\frac{h}{m v}$ )
for fourth orbit $n=4$
$\begin{gathered}
r=\frac{2 \lambda}{\pi} \\
\therefore \text { circumference }=2 \pi r=2 \pi \times \frac{2 \lambda}{\pi}=4 \lambda
\end{gathered}$
$\therefore \quad m v r=\frac{n h}{2 \pi}$
or
$r=\frac{n}{2 \pi} \cdot\left(\frac{h}{m v}\right) \text { or } r=\frac{n \lambda}{2 \pi}$
(From de -Broglie equation, $\lambda=\frac{h}{m v}$ )
for fourth orbit $n=4$
$\begin{gathered}
r=\frac{2 \lambda}{\pi} \\
\therefore \text { circumference }=2 \pi r=2 \pi \times \frac{2 \lambda}{\pi}=4 \lambda
\end{gathered}$
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