Question

In Bohr's atomic model, the potential energy of an electron at a position is $\frac{k{r}^2}{2}$ (where, $k$ is positive constant) then the quantized energy of the electron in $n^{th}$ orbit is

• A. $\frac{nh}{2\pi }\left(\frac{k}{m}\right)$
• B. $\frac{nh}{8\pi }{\left(\frac{k}{m}\right)}^{\frac{1}{2}}$
• C. $nh\left(\frac{k}{m}\right)$
• D. $\frac{nh}{\pi }{\left(\frac{k}{m}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is B $\frac{nh}{8\pi }{\left(\frac{k}{m}\right)}^{\frac{1}{2}}$

Given, $dU=\frac{k{r}^2}{2}$

Force acting on the electron is,

$\left|\begin{array}{c}{F}_e\end{array}\right|=\frac{dU}{dr}=\frac{d}{dr}\left(\frac{k{r}^2}{2}\right)=kr$

The orbital electrons always rotate with fixed (or) constant speed in a particular orbit,

$\Rightarrow F_e=\frac{m{v}^2}{r}$

$\Rightarrow kr=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt{\frac{k{r}^2}{m}}=r\sqrt{\frac{k}{m}}......\left(1\right)$

According to Bohr's theory, angular momentum is given by,

$mvr=\frac{nh}{2\pi }......\left(2\right)$

On putting the value of $v$ from eq $\left(1\right)$ in eq $\left(2\right)$ we get,

$\Rightarrow m\left(r\sqrt{\frac{k}{m}}\right)r=\frac{nh}{2\pi }$

$\Rightarrow m\sqrt{\frac{k}{m}}{r}^2=\frac{nh}{2\pi }$

$\Rightarrow \sqrt{km}{r}^2=\frac{nh}{2\pi }$

$\Rightarrow r=\sqrt{\frac{nh}{2\pi \sqrt{km}}}$

As we know, $(T.E{)}_n=\frac{(P.E{)}_n}{2}$

$\Rightarrow E_n=\frac{\frac{k{r}^2}{2}}{2}=\frac{k{r}^2}{4}$

$\Rightarrow E_n=\frac{k}{4}\left(\frac{nh}{\sqrt{km}2\pi }\right)$

$\Rightarrow E_n=\frac{nh}{8\pi }\sqrt{\frac{k}{m}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.

Why this question? Key idea: If the potential energy of an electron is given, then the force acting on the electron is calculated as, $F=-\frac{dU}{dr}$ Where, $U=$ Potential energy of the electron, $r=$distance from nucleus.