In Bohr's theory the potential of an electron at a position is $\frac{K r^2}{2}$, where $K$ is a constant. Then, the quantised energy of the electron in $n$th orbit is

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Aicharya · Accepted Answer

Given, potential energy of electron,
$\begin{aligned}
& & U=\frac{K r^2}{2} \\
\therefore & F & =\frac{d U}{d r}=\frac{d}{d r} \frac{K r^2}{2}=K r \\
\text {But } & F & =\frac{m v^2}{r} \Rightarrow K r=\frac{m v^2}{r} \\
\Rightarrow & v & =r \sqrt{\frac{K}{m}}
\end{aligned}$
According to Bohr's postulates,
$\begin{aligned}
L & =m v r=\frac{n h}{2 \pi} \\
\Rightarrow m r \cdot v & =\frac{n h}{2 \pi} \\
\Rightarrow m r \cdot r \sqrt{\frac{K}{m}} & =\frac{n h}{2 \pi} \text { [from Eq. (i)] } \\
\Rightarrow r^2 \sqrt{K m} & =\frac{n h}{2 \pi} \\
\Rightarrow r^2 & =\frac{n h}{2 \pi \sqrt{K m}}
\end{aligned}$
We know that, energy of electron in $n$th orbit,
$\begin{aligned}
B_n & =K r^2 \\
& =K \cdot \frac{n h}{2 \pi \sqrt{K m}} \text { [from Eq. (ii)] } \\
& =\frac{n h}{2 \pi} \sqrt{\frac{K}{m}}
\end{aligned}$

In Bohr's theory the potential of an electron at a position is $\frac{K r^2}{2}$, where $K$ is a constant.…

Solution