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  • The inverse square law in electrostatics is | F |=frac{e^{2}}{ (4 pi epsilon _{0} )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

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.

Answers (1)

We are given

\lambda=\frac{m_{p}c}{h}=\frac{m_{p}c^{2}}{hc}=\frac{(10^{2}m_{e})c^{2}}{hc}\\ =\frac{10^{-6}[0.51][1.6 \times 10^{-13}J]3\times 10^{8}ms^{-1}}{(1.05 \times 10^{-34}Js)(3 \times 10^{8}ms^{-1})}\\ =0.26\times 10^{7}m^{-1},\left [ \because m_{e}C^{2}=0.51MeV \right ]

\\r_{B}(\ Bohrs\ radius=0.51\AA=0.51 \times 10^{-10}m)\; \\or \lambda r_{B}=(0.26 \times 10^{7}m^{-1})(0.51 \times 10^{-10}m) =0.14 \times 10^{-13}<<1

Further\; as\; \left | F \right |=\left ( \frac{e^{2}}{4\pi \epsilon _{0}} \right )\left [ \frac{1}{r^{2}+\frac{\lambda}{r}} \right ]e^{-\lambda}r............(i)\\ 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}(e^{-\lambda r})\\ \frac{dz}{dr}=\left [\frac{1}{r}(e^{-\lambda r})(-\lambda)+(e^{-\lambda r})\left ( \frac{1}{r^{2}} \right ) \right ]\\ or\; dz=-\left [\left ( \frac{\lambda e^{-\lambda r}}{r}+\frac{ e^{-\lambda r}}{r^{2}} \right ) \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 ).......(ii)

we know that

mvr=h\Rightarrow v=\frac{h}{mr}; \; and \\ \frac{mv^{2}}{r}=F=\left ( \frac{e^{2}}{4 \pi \epsilon_{0} } \right )\left ( \frac{1}{r^{2}}+\frac{\lambda}{r} \right )\\\left [ putting\; e^{-\lambda r}\approx 1 \;in \; eqn.(i) \right ]\\ 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}}{mr^{3}}=\left ( \frac{e^{2}}{4 \pi \epsilon_{0} } \right )\left ( \frac{r+\lambda r^{2}}{r^{3}} \right )\\ or\;\frac{h^{3}}{mr^{3}}=\left ( \frac{e^{2}}{4 \pi \epsilon_{0} } \right )\left ( r+ \lambda r^{2} \right )...............(iii)\\ when \; \lambda=0, r={r_{B}}'\; and \\ \frac{h^{3}}{mr^{3}}=\left ( \frac{e^{2}}{4 \pi \epsilon_{0} } \right )r_{B}......................(iv)

from eq.(iii) and (iv)

r_{B}+r+\lambda r^{2}\\ let\; r=r+{B}+\delta \; so\;that\;from(iii)\\ r_{B}=(r_{B}+\delta)+\lambda(r_{B}^{2}+\delta^{2}+2\delta r_{B})\\ or 0=\lambda r_{B}^{2}+\delta(1+2 \lambda r_{B})\;(neglectinf \delta^{2})\\

or\; \delta =\frac{\lambda r_{B}^{2}}{(1+2 \lambda r_{B})}=(-\lambda r_{B}^{2})(1+\lambda r_{B}^{2})^{-1}\\ =(-\lambda r_{B}^{2})(1+\lambda r_{B}^{2})=\lambda r_{B}^{2} \;\;\left ( \because \lambda r_{B}^{2}<<1 \right )\\ from \;eq.(ii)\\ \mu_{i}=- \left ( \frac{e^{2}}{4 \pi \epsilon_{0} } \right )\frac{e^{-\lambda (r_{B}+\delta)}}{(r_{B}+\delta)}\\ =- \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} }\\ =-24.2eV

\left [\because e^{-\lambda (r_{B}+\delta)}\approx 1-\lambda(r_{B}-\delta) =1-\lambda r_{B} - \lambda \delta \approx 1- \lambda e_{B}\right ]\\ and \frac{1}{(r_{B}+\delta)}=\frac{1}{r_{B}(1+\delta/r_{B})}=\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{h^{2}}{m^{2}r^{2}} \right )\\ =\frac{h^{2}}{2mr^{2}}=\frac{h^{2}}{2m(r_{B}+\delta)^{2}}=\frac{h^{2}}{2mr_{B}^{2}+(1+\delta/r_{B})^{2}}\\ =\left ( \frac{h^{2}}{2mr_{B}^{2}} \right )\left ( 1+\frac{\delta}{r_{B}} \right )^{-2}=\left ( \frac{h^{2}}{2mr^{2}B} \right )\left ( 1-\frac{2\delta}{r_{B}} \right )\\ =(13.6)(1+2\lambda r_{B})eV \left ( as\;\frac{h^{2}}{2mr_{B}}=13.6 eV \; and \; \delta=-\lambda r_{B}^{2} \right )

The total energy of H-atom in the ground state=finalbenergy - initial energy

=(-13.6+27.2 \lambda r_{B})eV-(-13.6 eV)\\ =(27.2 \lambda r_{B})eV

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