The inverse square law in electrostatics is |F|= e^2 (4 _0)r^2 fo…

Question

The inverse square law in electrostatics is $|\text{F}|=\frac{\text{e}^2}{(4\pi\in_0)\text{r}^2}$ for the force between an electron and a proton. The $\Big(\frac{1}{\text{r}}\Big)$ dependence of |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 $|\text{F}|=\frac{\text{e}^2}{(4\pi\in_0)\text{r}^2}\Big[\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big].\text{e}\times\text{p}(-\lambda\text{r})$ where $\lambda=\frac{\text{m}_\text{p}\text{c}}{\text{h}}$ and $\text{h}=\frac{\text{h}}{2\pi}$. Estimate the change in the ground state energy of a H-atom if m_p were 10^-6 times the mass of an electron.

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Answer

We are given $\lambda=\frac{\text{m}_\text{p}\text{c}}{\text{h}}=\frac{\text{m}_\text{p}\text{c}^2}{\text{hc}}=\frac{(10^{-6}\text{m}_\text{e})\text{c}^2}{\text{hc}}$
$=\frac{10^{-6}[0.51](1.6\times10^{-13}\text{J})3\times10^8\text{ms}^{-1}}{(1.05\times10^{-34}\text{Js})(3\times10^8\text{ms}^{-1})}$
$=0.26\times10^7\text{m}^{-1},\ \ [\because\ \text{m}_\text{e}\text{c}^2=0.51\text{MeV}]$
$\text{r}_\text{B}(\text{Bohr's radius}=051\mathring{\text{A}}=0.51\times10^{-10}\text{m}$
or $\lambda\text{r}_\text{B}=(0.26\times10^7\text{m}^{-1})(0.51\times10^{-10}\text{m})$
$=0.14\times10^{-3}< < 1$
Furhter, as $|\text{F}|=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\bigg[\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\bigg]\text{e}^{\lambda\text{r}}\ .....(\text{i})$
and $|\text{F}|=\frac{\text{dU}}{\text{dr}},$
$\text{U}_\text{r}=\int|\text{F}|\text{dr}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\int\Big(\frac{\lambda_\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big)\text{dr}$
If $\text{z}=\frac{\text{e}^{-\lambda\text{r}}}{\text{r}}=\frac{1}{\text{r}}(\text{e}^{-\lambda\text{r}}),$
$\frac{\text{dz}}{\text{dr}}=\Big[\frac{1}{\text{r}}(\text{e}^{-\lambda\text{r}})(-\lambda)+(\text{e}^{-\lambda\text{r}})\Big(-\frac{1}{\text{r}^2}\Big)\Big]$
or $\text{dz}=-\Big[\frac{\lambda\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big]\text{dr}$
Thus $\int\Big(\frac{\lambda\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big)\text{dr}\ \Rightarrow-\int\text{dz}=-\text{z}=-\frac{\text{e}^{-\lambda\text{r}}}{\text{r}}$
$=-\Big(\frac{\text{e}^2}{4\pi\text{E}_0}\Big)\Big(\frac{\text{e}^{-\lambda\text{r}}}{\text{r}}\Big)\ .....(\text{ii})$
We know that
$\text{mvr}=\text{h}\Rightarrow\ \text{v}=\frac{\text{h}}{\text{mr}};\text{ and }$
$\frac{\text{mv}^2}{\text{r}}=\text{F}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\Big(\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big)$
[Putting $\text{e}^{-\lambda\text{r}}\approx1$ in eqn. (i)]
Thus, $\Big(\frac{\text{m}}{\text{r}}\Big)\Big(\frac{\text{h}^2}{\text{m}^2\text{r}^2}\Big)=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\Big(\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big)$
or $\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\Big(\frac{\text{r}+\lambda\text{r}^2}{\text{r}^3}\Big)$
or $\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)({\text{r}+\lambda\text{r}^2})\ .....(\text{iii})$
When $\lambda=0,\text{r}=\text{r}'_\text{B}$ and
$\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\text{r}_\text{B}\ .....(\text{iv})$
From eqns. (iii) and (iv),
$\text{r}_\text{B}+\text{r}+\lambda\text{r}^2$
Let $\text{r}=\text{r}_\text{B}+\delta$ so that from (iii)
$\text{r}_\text{B}=(\text{r}_\text{B}+\delta)+\lambda(\text{r}^2_\text{B}+\delta^2+2\delta\text{r}_\text{B})$
or $0=\lambda\text{r}_\text{B}^2+\delta(1+2\lambda\text{r}_\text{B})\ \ (\text{neglecting }\delta^2)$
or $\delta=-\frac{\lambda\text{r}_\text{B}^2}{(1+2\lambda\text{r}_\text{B})}=(-\lambda\text{r}_\text{B}^2)(1+2\lambda_\text{B})^{-1}$
$=(-\lambda\text{r}_\text{B}^2)(1+2\lambda_\text{B})^{-1}=-\lambda\text{r}_\text{B}^2\ \ (\because\ \lambda\text{r}_\text{B}< < 1)$
From eqn. (ii) $\text{u}_\text{r}=-\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\frac{\text{e}^{-\lambda(\text{r}_\text{B}+\delta)}}{(\text{r}_\text{B}+\delta)}$
$=-\Big(\frac{\text{e}^2}{4\pi\in_0}\frac{1}{\text{r}_\text{B}}\Big)\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)(1-\lambda\text{r}_\text{B})\approx-\frac{\text{e}^2}{4\pi\in_0\text{r}_\text{B}}$
$=-24.2\text{eV}$
$[\because\ \text{e}^{-\lambda(\text{r}_\text{B}+\delta)}\approx\ 1-\lambda(\text{r}_\text{B}+\delta)=1-\lambda\text{r}_\text{B}-\lambda\delta\approx1-\lambda\text{e}_\text{B}]$
and $\frac{1}{(\text{r}_\text{B}+\delta)}=\frac{1}{\text{r}_\text{B}(1+\frac{\delta}{\text{r}_\text{B}})}=\frac{1}{\text{r}_\text{B}}\Big(1+\frac{\delta}{\text{r}_\text{B}}\Big)^{-1}$
$=\frac{1}{\text{r}_\text{B}}\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)$
Further, KE of the electron,
$\text{K}=\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{m}\Big(\frac{\text{h}^2}{\text{m}^2\text{r}^2}\Big)$
$=\frac{\text{h}^2}{2\text{mr}^2}=\frac{\text{h}^2}{2\text{m}(\text{r}_\text{B}+\delta)^2}=\frac{\text{h}^2}{2\text{mr}_\text{B}^2+(1+\frac{\delta}{\text{r}_\text{B}})^2}$
$\Big(\frac{\text{h}^2}{2\text{mr}^2_\text{B}}\Big)\Big(1+\frac{\delta}{\text{r}_\text{B}}\Big)^{-2}=\Big(\frac{\text{h}^2}{2\text{mr}^2_\text{B}}\Big)\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)$
$=(13.6)(1+2\lambda\text{r}_\text{B})\text{eV}$
$\Big(\text{as}\frac{\text{h}^2}{2\text{mr}^2_\text{B}}=13.6\text{eV and }\delta=-\lambda\text{r}^2_\text{B}\Big)$
Total energy of H-atom in the ground state = final energy - initial energy
$=(-13.6+27.2\lambda\text{r}_\text{B})\text{eV}-(-13.6\text{eV})$
$=(27.2\lambda\text{r}_\text{B})\text{eV}$

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