Question

Suppose that the potential energy of an hypothetical atom consisting of a proton and an electron is given by $U=-k{e}^2/3r^3$. Then if Bohr's postulates are applied to this atom, then the radius of the $nt$ orbit will be proportional to:-

• A. $n^2$
• B. $1/n^2$
• C. $n^3$
• D. $1/n^3$

Answer 1

The correct option is D $1/n^3$

$\begin{array}{c}\text{Given,}\\ \text{(potential Energy)}=\frac{-k{e}^2}{3r^3}\\ \text{we know relation between force and}\\ \text{potential Energy}\end{array}$

$\begin{array}{c}F=\frac{-dy}{dr}\\ F=-\frac{d}{dr}\left(\frac{-K{e}^2}{3r^3}\right)\\ F=\frac{K{e}^2}{r^4}\end{array}$

$\begin{array}{c}\text{If Bohr's postulate is true for this}\\ \text{Force}=\text{centripital force}=\frac{m{v}^2}{r}\end{array}$

$\text{So,}\frac{m{v}^2}{r}=\frac{k{e}^2}{r^4}-\left(i\right)$

$\begin{array}{c}\text{and from Bohr's postulate}\\ \text{angular momentum}\left(mvr\right)=\frac{nh}{2\pi }\end{array}$

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

$\text{Putting value of v from (ii) in (i)}$

$\frac{m{\left(\frac{nh}{2\pi mr}\right)}^2}{r}=\frac{k{e}^2}{r^4}$

$\frac{n^2{h}^2}{4{\pi }^{2}mr}=\frac{k{e}^2}{r^4}$

$r\alpha \frac{1}{n^2}$