Question:28
The inverse square law in electrostatics is $\left | F \right |=\frac{e^{2}}{\left (4 \pi \epsilon _{0} \right )r^{2}}$ for the force between an electron and a proton. The $\frac{1}{r}$ 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 mp, force would be modified to $\left | F \right |=\frac{e^{2}}{\left (4 \pi \epsilon _{0} \right )r^{2}}\left [ \frac{1}{r^{2}} +\frac{\lambda}{r}\right ].\exp \left ( -\lambda r \right )$ where $\lambda=m_{p}c/h$ and $h=\frac{h}{2\pi}$ .
Estimate the change in the ground state energy of a H-atom if mp were 10–6 times the mass of an electron.
Answer:
We are given
$\begin{aligned}
& \lambda=\frac{m_p c}{h}=\frac{m_p c^2}{h c}=\frac{\left(10^2 m_e\right) c^2}{h c} \\
&= \frac{10^{-6}[0.51]\left[1.6 \times 10^{-13} \mathrm{~J}\right] 3 \times 10^8 \mathrm{~ms}^{-1}}{\left(1.05 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)} \\
&= 0.26 \times 10^7 \mathrm{~m}^{-1},\left[\because m_e C^2=0.51 \mathrm{MeV}\right] \\
& r_B\left(\text { Bohrs radius }=0.51 A=0.51 \times 10^{-10} \mathrm{~m}\right) \\
& \text { or } \lambda r_B=\left(0.26 \times 10^7 \mathrm{~m}^{-1}\right)\left(0.51 \times 10^{-10} \mathrm{~m}\right)=0.14 \times 10^{-13} \ll 1
\end{aligned}$
Further as $|F|=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left[\frac{1}{r^2+\frac{\lambda}{r}}\right] e^{-\lambda} r\ldots\ldots(i)$
and $|F|=\frac{d U}{d r}$
$\begin{aligned}
& U_r=\int|F| d r=\left(\frac{e^2}{4 \pi \epsilon_0}\right) \int\left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right) d r \\
& I f z=\frac{e^{-\lambda r}}{r}=\frac{1}{r}\left(e^{-\lambda r}\right) \\
& \frac{d z}{d r}=\left[\frac{1}{r}\left(e^{-\lambda r}\right)(-\lambda)+\left(e^{-\lambda r}\right)\left(\frac{1}{r^2}\right)\right]
\end{aligned}$
$\text { or } d z=-\left[\left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right)\right] d r$
Thus $\int\left(\frac{\lambda e^{-\lambda r}}{r}+\frac{e^{-\lambda r}}{r^2}\right) d r \Rightarrow-\int d z=-z=-\frac{e^{\lambda r}}{r}$ $=-\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{e^{\lambda r}}{r}\right) \ldots\ldots(ii)$
We know that,
$\begin{gathered}
m v \tau=h \Rightarrow v=\frac{h}{m m} ; \text { and } \\
\frac{m v^2}{r}=F=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{1}{r^2}+\frac{\lambda}{r}\right)
\end{gathered}$
$\left[\right.$ putting $e^{-\lambda r} \approx 1$ in eqn.(i)]
Thus $\left(\frac{m}{r}\right)\left(\frac{h^2}{4 \pi \epsilon_0}\right)=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{1}{r^2}+\frac{\lambda}{r}\right)$
or, $\frac{h^3}{m r^3}=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(\frac{r+\lambda r^2}{r^3}\right)$
or, $\frac{h^3}{m r^3}=\left(\frac{e^2}{4 \pi \epsilon_0}\right)\left(r+\lambda r^2\right) \ldots\ldots(iii)$
when $\lambda=0, r=r_B^{\prime}$ and
$\frac{h^3}{m r^3}=\left(\frac{e^2}{4 \pi \epsilon_0}\right) r_B \ldots\ldots(iv)$
from eq.(iii) and (iv)
$\begin{aligned}
& r_B+r+\lambda r^2 \\
& \text { let } r=r+B+\delta \text { so that from }(i i i) \\
& r_B=\left(r_B+\delta\right)+\lambda\left(r_B^2+\delta^2+2 \delta r_B\right) \\
& \text { or } 0=\lambda r_B^2+\delta\left(1+2 \lambda r_B\right)\left(\text { neglectin } f \delta^2\right) \\
& \text { or } \delta=\frac{\lambda r_B^2}{\left(1+2 \lambda r_B\right)}=\left(-\lambda r_B^2\right)\left(1+\lambda r_B^2\right)^{-1} \\
& =\left(-\lambda r_B^2\right)\left(1+\lambda r_B^2\right)=\lambda r_B^2 \quad\left(\because \lambda r_B^2 \ll 1\right)
\end{aligned}$
from eg.(ii)
$\begin{aligned}
\mu_i&=-\left(\frac{e^2}{4 \pi \epsilon_0}\right) \frac{e^{-\lambda\left(r_B+\delta\right)}}{\left(r_B+\delta\right)} \\
& =-\left(\frac{e^2}{4 \pi \epsilon_0} \frac{1}{r_B}\right)\left(1-\frac{\delta}{r_B}\right)\left(1-\lambda r_B\right)\\
& =-\frac{e^2}{4 \pi \epsilon_0 r_B} \\
& =-24.2 e V
\end{aligned}$
$\left[\because e^{-\lambda\left(r_B+\bar{\sigma}\right)} \approx 1-\lambda\left(r_B-\delta\right)=1-\lambda r_B-\lambda \delta \approx 1-\lambda e_B\right]$
$\text { and } \frac{1}{\left(r_B+\delta\right)}=\frac{1}{r_B\left(1+\delta / r_B\right)}=\frac{1}{r_B}\left(1+\frac{\delta}{r_B}\right)^{-1}$
$=\frac{1}{r_B}\left(1-\frac{\delta}{r_B}\right)$
Further KE of the electron
$\begin{aligned}
K=&\frac{1}{2} m v^2=\frac{1}{2} m\left(\frac{h^2}{m^2 r^2}\right) \\
= & \frac{h^2}{2 m r^2}=\frac{h^2}{2 m\left(r_B+\delta\right)^2}=\frac{h^2}{2 m r_B^2+\left(1+\delta / r_B\right)^2} \\
= & \left(\frac{h^2}{2 m r_B^2}\right)\left(1+\frac{\delta}{r_B}\right)^{-2}=\left(\frac{h^2}{2 m r^2 B}\right)\left(1-\frac{2 \delta}{r_B}\right) \\
= & (13.6)\left(1+2 \lambda r_B\right) e V\left(a s \frac{h^2}{2 m r_B}=13.6 e V \text { and } \delta=-\lambda r_B^2\right)
\end{aligned}$
The total energy of H -atom in the ground state-final energy - initial energy
$\begin{aligned}
& =\left(-13.6+27.2 \lambda r_B\right) \mathrm{eV}-(-13.6 \mathrm{cV}) \\
& = \left(27.2 \lambda_{r_B}\right) \mathrm{eV}
\end{aligned}$
