Question

The inverse square law in electrostatics is$\left|\vec{F}\right|=\frac{e^2}{\left(4\pi {\epsilon }_0\right)\cdot r^2}$ for the force between an electron and a proton. The $(\frac{1}{r})$ dependence of $|\vec{F}|$ can be understood in quantum theory as being due to the fact that the ‘particle’ of light (photon) is massless. If photons had a mass $m_p$, force would be modified to
$\left|\vec{F}\right|=\frac{e^2}{\left(4\pi {\epsilon }_0\right)r^2}\left[\frac{1}{r^2}+\frac{\lambda }{r}\right]\cdot exp\left(-\lambda r\right)$
where $\lambda =m_{p}c/\hslash$ and $\hslash =\frac{h}{2\pi }$ The change in the ground state energy $(eV)$ of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron. ($r_B=$ Bohr’s radius)

• A. $18.6\lambda r_B$
• B. $-27$
• C. $27.2\lambda r_B$
• D. $-\lambda r_B$

Answer 1

Answer: (C)

As $\lambda =\frac{m_{p}c}{\hslash }=\frac{m_p{c}^2}{\hslash c}$
$=\frac{\left(10^{-6}{m}_e\right)c^2}{\hslash c}$
$=\frac{10^{-6}\left[0.51\right]\left(1.6\times 10^{-16}J\right)}{\left(1.05\times 10^{-34}Js\right)\left(3\times 10^{8}m{s}^{-1}\right)}$
$=0.26\times 10^7{m}^{-1}\left[\because m_e{c}^2=0.51MeV\right]$
$r_B$ (Bohr’s radius) $=0.51\mathring{A}=0.51\times 10^{-10}m$
or $\lambda r_B=\left(0.26\times 10^7{m}^{-1}\right)\left(0.51\times 10^{-10}m\right)$
$=0.14\times 10^{-3}<<1$
Further, as $|F|=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\left[\frac{1}{r^2}+\frac{\lambda }{r}\right]e^{-\lambda r}...\left(i\right)$
and $\left|F\right|=\frac{dU}{dr}$,
$U_r=\int \left|F\right|dr=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\int \left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right)dr$
If $z=\frac{e^{-\lambda r}}{r}=\frac{1}{r}\left(e^{-\lambda r}\right)$,
$\frac{dz}{dr}=\left[\frac{1}{r}\left(e^{-\lambda r}\right)\left(-\lambda \right)+\left(e^{-\lambda r}\right)\left(-\frac{1}{r^2}\right)\right]$
or $dz=-\left[\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right]dr$
Thus $\int \left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right)dr$
$\Rightarrow -\int dz=-z=-\frac{e^{-\lambda r}}{r}$
$=-\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\left(\frac{e^{-\lambda r}}{r}\right)...\left(ii\right)$
We know that, $mvr=\hslash$
$\Rightarrow v=\frac{\hslash }{mr};$ and
$\frac{m{v}^2}{r}=F=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\left(\frac{1}{r^2}+\frac{\lambda }{r}\right)$
$\left[\text{putting}{e}^{-\lambda r}\approx 1\text{in eqn}.\left(i\right)\right]$
Thus $\left(\frac{m}{r}\right)\left(\frac{{\hslash }^2}{m^2{r}^2}\right)=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\left(\frac{1}{r^2}+\frac{\lambda }{r}\right)$
or $\frac{{\hslash }^2}{m{r}^3}=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\left(\frac{r+\lambda r^2}{r^3}\right)$
or $\frac{{\hslash }^3}{m}=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\left(r+\lambda r^2\right)...\left(iii\right)$
When $\lambda =0,r=r_B$, and $\frac{{\hslash }^2}{m}=\left(\frac{e^2}{4\pi {\epsilon }_0}\right)r_B...\left(iv\right)$
From eqns. $\left(iii\right)$ and $\left(iv\right),r_B=r+\lambda r^2$
Let $r=r_B+\delta$ so that from $\left(iii\right)$
$r_B=\left(r_B+\delta \right)+\lambda \left(r_B^2+{\delta }^2+2\delta r_B\right)$
or $0=\lambda r_B^2+\delta \left(1+2\lambda r_B\right)$(neglecting ${\delta }^2)$
or $\delta =-\frac{\lambda r_B^2}{\left(1+2\lambda r_B\right)}=\left(-\lambda r_B^2\right){\left(1+2\lambda r_B\right)}^{-1}$
$=\left(-\lambda r_B^2\right)\left(1-2\lambda r_B\right)=-\lambda r_B^2\left(\because \lambda r_B<<1\right)$
From eqn. $\left(ii\right)U_r=-\left(\frac{e^2}{4\pi {\epsilon }_0}\right)\frac{e^{-\lambda \left(r_B+\delta \right)}}{\left(r_B+\delta \right)}$
$=-\left(\frac{e^2}{4\pi {\epsilon }_0}\frac{1}{r_B}\right)\left(1-\frac{\delta }{r_B}\right)\left(1-\lambda r_B\right)\approx -\frac{e^2}{4\pi {\epsilon }_0{r}_B}$
$=-27.2eV$
$\left[\because e^{-\lambda \left(r_B+\delta \right)}\approx 1-\lambda \left(r_B+\delta \right)=1-\lambda r_B-\lambda \delta \approx 1-\lambda r_B\right]$
and $\frac{1}{\left(r_B+\delta \right)}=\frac{1}{r_B\left(1+\delta /r_B\right)}$
$=\frac{1}{r_B}{\left(1+\frac{\delta }{r_B}\right)}^{-1}$
$=\frac{1}{r_B}\left(1-\frac{\delta }{r_B}\right)$
Further, KE of the electron, $K=\frac{1}{2}m{v}^2$
$=\frac{1}{2}m\left(\frac{{\hslash }^2}{m^2{r}^2}\right)$
$=\frac{{\hslash }^2}{2m{r}^2}=\frac{{\hslash }^2}{2m{\left(r_B+\delta \right)}^2}$
$=\frac{{\hslash }^2}{2m{r}_B^2+\left(1+\delta r_B^2\right)}$
$=\left(\frac{{\hslash }^2}{2m{r}_B^2}\right){\left(1+\frac{\delta }{r_B}\right)}^{-2}$
$=\left(\frac{{\hslash }^2}{2m{r}_B^2}\right)\left(1-\frac{2\delta }{r_B}\right)$
$=\left(13.6\right)\left(1+2\lambda r_B\right)eV$
(as $\frac{h^2}{2m{r}_B^2}=13.6eV$ and $\delta =-\lambda r_B^2)$
Total energy of $H$-atom in the ground state
$=$ final energy - initial energy
$=(13.6+27.2\lambda r_B)eV-(-13.6eV)$
$=(27.2\lambda r_B)eV$