The inverse square law in electrostatic is $|F|=\frac{\mathrm{e}^2}{\left(4 \pi \varepsilon_0\right) \mathrm{r}^2}$ for the force between an electron and a proton. The $\left(\frac{1}{r}\right)$ dependence of $|\mathbf{F}|$ 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
$|\mathrm{F}|=\frac{\mathrm{e}^2}{\left(4 \pi \varepsilon_0\right) \mathrm{r}^2}\left[\frac{1}{\mathrm{r}^2}+\frac{\lambda}{\mathrm{r}}\right] \cdot \exp (-\lambda \mathrm{r})$ where $\lambda=\frac{\mathrm{m}_{\mathrm{p}} \mathrm{c}}{\hbar}$ and $\hbar=\frac{\mathrm{h}}{2 \pi}$. Estimate the change in the ground state energy of a $\mathrm{H}$-atom if $\mathrm{m}_{\mathrm{p}}$ were $10^{-6}$ times the mass of an electron.

Options:

None

Solution:
As given that, $m_p=10^{-6}$ times, the mass of an electron, the energy associated with it is given by
$$
\begin{aligned}
\mathrm{E} &=\mathrm{m}_{\mathrm{p}} \mathrm{c}^2=10^{-6} \times \text { electron mass } \times \mathrm{c}^2 \\
& \approx 10^{-6} \times 0.5 \mathrm{MeV} \\
& \approx 10^{-6} \times 0.5 \times 1.6 \times 10^{-13} \\
\mathrm{E} & \approx 0.8 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
The wavelength associated with is given by
$$
\begin{aligned}
&\lambda=\frac{\hbar}{\mathrm{m}_{\mathrm{p}} \mathrm{c}}=\frac{\hbar \mathrm{c}}{\mathrm{m}_{\mathrm{p}} \mathrm{c}^2}=\frac{10^{-34} \times 3 \times 10^8 \times 6.62}{0.8 \times 10^{-19}} \\
&\approx 2.4 \times 10^{-7} \mathrm{~m} \gg \text { Bohr radius } \mathrm{r}_3 \\
&|\mathrm{~F}|=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0}\left[\frac{1}{\mathrm{r}^2}+\frac{\lambda}{\mathrm{r}}\right] \exp (-\lambda r) \\
&\text { where, } \lambda^{-1}=\frac{\hbar}{m_p \mathrm{c}} \approx 4 \times 10^{-7} \mathrm{~m} \gg>\mathrm{r}_{\mathrm{B}} \\
&\therefore \quad\left(\lambda \ll < \frac{1}{\mathrm{r}_{\mathrm{B}}} \text { i.e., } \lambda \mathrm{r}_{\mathrm{B}} \ll < 1\right) \\
&U(r)=-\frac{e^2}{4 \pi \varepsilon_0} \cdot \frac{\exp (-\lambda r)}{r} \\
&
\end{aligned}
$$
$$
\begin{aligned}
&\therefore \quad \mathrm{mvr}=\hbar \quad \therefore \mathrm{v}=\frac{\hbar}{\mathrm{mr}} \\
&\text { Also, } \frac{\mathrm{mv}^2}{\mathrm{r}}=\left(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0}\right)\left[\frac{1}{\mathrm{r}^2}+\frac{\lambda}{\mathrm{r}}\right] \\
&\therefore \quad \frac{\hbar^2}{\mathrm{mr}^3}=\left(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0}\right)\left[\frac{1}{\mathrm{r}^2}+\frac{\lambda}{\mathrm{r}}\right] \\
&\therefore \quad \frac{\hbar^2}{\mathrm{~m}}=\left(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0}\right)\left[\mathrm{r}+\pi \mathrm{r}^2\right] \\
&\text { If } \lambda=0 ; \mathrm{r}=\mathrm{r}_{\mathrm{B}}=\frac{\hbar}{\mathrm{m}} \cdot \frac{4 \pi \varepsilon_0}{\mathrm{e}^2} \\
&\frac{\hbar^2}{\mathrm{~m}}=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0} \cdot \mathrm{r}_{\mathrm{B}}
\end{aligned}
$$
Since, $\lambda^{-1} \gg>r_B$, put $r=r_B+\delta$
$$
\begin{aligned}
&r_B=r+\lambda r^2 \\
&r_B=\left(r_B+\delta\right)+\lambda\left(r_B+\delta\right)^2 \\
&\therefore \quad r_B=r_B+\delta+1\left(r_B^2+\delta^2+2 \delta r_B\right) ; \text { neglect } \delta^2
\end{aligned}
$$
or $0=\lambda r_B^2+\delta\left(1+2 \lambda r_B\right)$
$$
\delta=\frac{-\lambda \mathrm{r}_{\mathrm{B}}^2}{1+2 \lambda \mathrm{r}_{\mathrm{B}}} \approx \lambda \mathrm{r}_{\mathrm{B}}^2\left(1-2 \lambda \mathrm{r}_{\mathrm{B}}\right)
$$
Since, $\left(\lambda r_B \ll < 1\right)$ is very small
So, $\delta=-\lambda r_B^2$
$$
\begin{aligned}
\mathrm{V}(\mathrm{r}) &=-\frac{\mathrm{e}^2}{4 \pi \varepsilon_0} \cdot \frac{\exp \left(-\lambda \delta-\lambda \mathrm{r}_{\mathrm{B}}\right)}{\left(\mathrm{r}_{\mathrm{B}}+\delta\right)} \\
&=\frac{-\mathrm{e}^2}{4 \pi \varepsilon_0} \frac{\mathrm{e}^{-\lambda\left(\delta+\mathrm{r}_{\mathrm{B}}\right)}}{\mathrm{r}_{\mathrm{B}}}\left[1+\frac{\delta}{\mathrm{r}_{\mathrm{B}}}\right]
\end{aligned}
$$
$$
\begin{aligned}
&\left(\because r=r_B+\delta\right) \\
&\mathrm{V}(\mathrm{r})=-\frac{\mathrm{e}^2 \mathrm{e}^{-\lambda \mathrm{r}}}{4 \pi \varepsilon_0} \frac{1}{\mathrm{r}_{\mathrm{B}}}\left[\left(1-\frac{\delta}{\mathrm{r}_{\mathrm{B}}}\right)\right] \\
&\mathrm{V}(\mathrm{r}) \cong(-27.2 \mathrm{eV}) \text { remains unchanged } \\
&\mathrm{KE}=-\frac{1}{2} \mathrm{mv}^2=\frac{-1}{2} \mathrm{~m} \cdot \frac{\mathrm{h}^2}{\mathrm{~m}^2 \mathrm{r}^2} \\
&=\frac{-\mathrm{h}^2}{2\left(\mathrm{r}_{\mathrm{B}}+\delta\right)^2 \mathrm{~m}}=\frac{\mathrm{h}^2}{2 \mathrm{r}_{\mathrm{B}}^2 \mathrm{~m}}\left(1-\frac{2 \delta}{\mathrm{r}_{\mathrm{B}}}\right) \\
&\left(\because \mathrm{v}=\frac{\mathrm{h}}{\mathrm{mr}}\right) \\
&=(13.6 \mathrm{eV})\left[1+2 \lambda \mathrm{r}_{\mathrm{B}}\right] \\
&\text { Total energy } \\
&(\mathrm{T} . \mathrm{E})=-\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}_{\mathrm{B}}}+\frac{\mathrm{h}^2}{2 \mathrm{r}_{\mathrm{B}}^2 \mathrm{~m}}\left[1+2 \lambda \mathrm{r}_{\mathrm{B}}\right] \\
&
\end{aligned}
$$
$$
\begin{aligned}
&=-27.2+13.6\left[1+2 \lambda r_B\right] \mathrm{eV} \\
&=\left[-27.2+13.6+27.2 \lambda r_B\right] e V \\
&(\text { T.E. })=-13.6+27.2 \lambda r_B \mathrm{eV}
\end{aligned}
$$
Change in energy
$$
=-13.6+27.2 \lambda r_B-(-13.6)=27.2 \lambda r_B e V
$$

Topic: Physics, Atomic Physics