Question

The frequency of revolution of electron in $n^{th}$ Bohr orbit is $V_n$. The graph between $logn$ and $log\left(\frac{V_n}{V_1}\right)$ may be

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We know that
$v\propto \frac{Z}{n}$
and $r\propto \frac{n^2}{Z}$
Hence $v$ and $r$ can be written as
$v=c_1\frac{Z}{n}$ and $r=c_2\frac{n^2}{Z}$, where both $c_1$ and $c_2$ are constants.

We know that Frequency $V_n=\frac{1}{T}=\frac{v}{2\pi r}$
Hence $V$ can be written as $V_n=c{n}^{-3}$

Now, $\frac{V_n}{V_1}=n^{-3}$
$\therefore log\frac{V_n}{V_1}=-3logn$

Hence graph between $logn$ and $log\left(\frac{V_n}{V_1}\right)$ will be a linear graph with negative slope passing through origin.