Question

The inverse square law in electrostatics is$\left|\vec{F}\right| = \frac{e^{2}}{\left(4\pi\varepsilon_{0}\right)\cdot r^{2}}$ 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
$\left|\vec{F}\right| = \frac{e^{2}}{\left(4\pi\varepsilon_{0}\right)r^{2}}\left[\frac{1}{r^{2}} + \frac{\lambda}{r}\right] \cdot exp \left(-\lambda r\right)$
where $\lambda = m_pc/\hbar$ and $\hbar= \frac{h}{2\pi}$ 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}{\hbar} = \frac{m_{p}c^{2}}{\hbar c} $
$= \frac{\left(10^{-6}m_{e}\right)c^{2}}{\hbar c} $
$ =\frac{ 10^{-6}\left[0.51\right]\left(1.6\times10^{-16} J\right)}{\left(1.05\times10^{-34} J s\right)\left(3\times10^{8} ms^{-1}\right)}$
$ = 0.26 \times10^{7} m^{-1}\left[\because m_{e}c^{2} = 0.51 MeV\right] $
$ r_{B} $ (Bohr’s radius) $ = 0.51 \mathring{A} = 0.51 \times10^{-10} m$
or $\lambda r_{B} = \left(0.26 \times10^{7} m^{-1}\right)\left(0.51\times10^{-10}m\right) $
$= 0.14 \times10^{-3} < < 1$
Further, as $|F| = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left[\frac{1}{r^{2}} +\frac{\lambda}{r}\right]e^{-\lambda r}...\left(i\right) $
and $\left|F\right| = \frac{dU}{dr}$,
$U_{r} = \int\left|F\right|dr = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\int\left(\frac{\lambda e^{-\lambda r}}{r} + \frac{e^{-\lambda r}}{r^{2}}\right)dr$
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\varepsilon_{0}}\right)\left(\frac{e^{-\lambda r}}{r}\right)...\left(ii\right) $
We know that, $mvr =\hbar $
$\Rightarrow v= \frac{\hbar}{mr};$ and
$\frac{mv^{2}}{r} = F = \left(\frac{e^{2}}{4\pi\varepsilon_{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{\hbar^{2}}{m^{2}r^{2}}\right) = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(\frac{1}{r^{2}} + \frac{\lambda}{r}\right) $
or $\frac{\hbar^{2}}{mr^{3} } = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(\frac{r+\lambda r^{2}}{r^{3}}\right)$
or $\frac{\hbar^{3}}{m} = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)\left(r+\lambda r^{2}\right)...\left(iii\right)$
When $\lambda = 0, r= r_{B}$, and $\frac{\hbar^{2}}{m} = \left(\frac{e^{2}}{4\pi\varepsilon_{0}}\right)r_{B} ...\left(iv\right)$
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\varepsilon_{0}}\right) \frac{e^{-\lambda\left(r_{B}+\delta\right)}}{\left(r_{B} +\delta\right)} $
$ = -\left(\frac{e^{2}}{4\pi\varepsilon_{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\varepsilon_{0} r_{B}}$
$= -27.2 eV$
$ \left[\because e^{-\lambda\left(r_{B}+\delta\right)} \approx1 -\lambda \left(r_{B} +\delta\right) = 1 -\lambda r_{B} -\lambda\delta \approx1-\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, $K= \frac{1}{2}mv^{2}$
$= \frac{1}{2}m\left(\frac{\hbar^{2}}{m^{2}r^{2}}\right)$
$ = \frac{\hbar^{2}}{2mr^{2}}= \frac{\hbar^{2}}{2m\left(r_{B}+\delta\right)^{2}} $
$ = \frac{\hbar^{2}}{2mr_{B}^{2} +\left(1+\delta r_{B}^{2}\right)} $
$ = \left(\frac{\hbar^{2}}{2mr_{B}^{2}}\right)\left(1+\frac{\delta}{r_{B}}\right)^{-2}$
$=\left(\frac{\hbar^{2}}{2mr_{B}^{2}}\right)\left(1-\frac{2\delta}{r_{B}}\right)$
$ = \left(13.6\right)\left(1+2\lambda r_{B}\right)eV $
(as $\frac{h^{2}}{2mr_{B}^{2}} = 13.6 eV$ and $\delta = -\lambda r_{B}^{2}) $
Total energy of $H$-atom in the ground state
$=$ final energy - initial energy
$= (13.6 + 27.2 \lambda r_B) \,eV - (-13.6 \,eV)$
$= (27.2 \lambda r_B) \,eV$