Question

Suppose the potential energy between electron and proton at a distance $r$ is given by $\frac{-K{e}^2}{3r^3}$ Use Bohr's theory to find velocity and its value will be :

• A. $V=\frac{k{e}^{2}2\pi }{r^{3}nh}$
• B. $V=\frac{n{h}^2}{2\pi mr}$
• C. $V=\frac{n^{2}h}{2\pi mr}$
• D. Can't be found

Answer 1

Answer: (A)

$V=\frac{k{e}^{2}2\pi }{r^{3}nh}$

$\begin{array}{c}\text{Given,}\\ \text{Potential Energy}\left(U\right)=-\frac{K{e}^2}{3r^3}\end{array}$

$\begin{array}{c}\text{We know,}\\ \begin{array}{cc}\text{Force}& =-\frac{du}{dr}\\ F& =-\frac{d}{dr}\left(\frac{-k{e}^2}{3r^3}\right)\\ F& =\frac{-K{e}^2}{r^4}\end{array}\end{array}$

$\begin{array}{c}\text{By Bohr's theory force should equal to}\\ \text{centripetal force}\end{array}$

$\begin{array}{cc}\frac{k{e}^2}{r^4}& =\frac{m_e{v}^2}{r}\\ m_e{v}^2& =\frac{k{e}^2}{r^3}-\left(i\right)\end{array}$

$\begin{array}{c}\text{Now, from conservation if angular. momentam}\\ m_{e}vr=\frac{nh}{2\pi }-\left(ii\right)\end{array}$

$\begin{array}{c}\text{Now,}\\ \text{divide equation}\left(i\right)÷\left(ii\right)\\ v=\frac{k{e}^2\cdot 2\pi }{r^{3}nh}\\ \end{array}$