Question

The inverse square law in electrostatics is |F| = `e^2/((4πε_0).r^2)` for the force between an electron and a proton. The `(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 m_p, force would be modified to |F| = `e^2/((4πε_0)r^2) [1/r^2 + λ/r]`, exp (– λr) where λ = m_pc/h and h = `h/(2π)`. Estimate the change in the ground state energy of a H-atom if m_p were 10^-6 times the mass of an electron.

Answer 1

Mass of photon = 9.1 × 10^–31 × 10^–6 kg

= 9.1 × 10^–37 kg

Wavelength associated with a photon = `h/(m_pc)`

λ = `(6.62 xx 10^-34)/(9.1 xx 10^-37 xx 3 xx 10^8)`

= `(6.62)/(9.1 xx 3) xx 10^(-34+37-8) - 2.4 xx 10^-7` >>  r_A  (see Q.26)

λ << `1/r_A` < `e λ_(r_A)` << 1

`U(r) = (- e^2e^(-λ))/(πε_0r)`

`mvr = h/(2pi) = h` or `v = h/(mr)`  .....(I)

`(mv^2)/v e^2/(4πε_0) [1/r^2 + λ/r]` .....`[∵ F = e^2/((4πε_0)r^2) [1/r^2 + λ/r] e^(-λr) "given"]`

`m/r * h^2/(m^2r^2) = e^2/(4πε_0) [(1 + λr)/r^2]`

`h^2/(mr) = e^2/(4πε_0) (1 + λr)`

`(h^2 4πε_0)/(me^2) = (r + λr^2)`

If λ = 0, `(h^2 4πε_0)/(me^2) = (r + λr^2)`  .....[Neglecting r²]

`h/m = e^2/(4πε_0) r_A`  .....`(r_A = r + λr^2)`

∵ λ_A >>> r_B and r = r_a + δ

Taking r_A + r + λr²  .....[∵ r = (r_a + δ]

r_A = (r_A + δ) + λ(r_A + δ) ....(Put r = r_A + δ²)

0 = `δ + λr_A^2 + 2r_A δλ`  .....(Neglecting small term δ²)

0 = `δ + 2r_A δλ + λr_A^2`

⇒ `δ[1 + 2r_A λ] = - λr_A^2`

δ = `(-λr_A^2)/((1 + 2r_Aλ)) = - λr_A^2 (1 + 2r_Aλ)^-1`

δ = `-λr_A^2 [1 - 2r_Aλ] = - λr_A^2 + 2r_A^3λ^2`

∴ λ and r_A << 1 so `r_A^3λ^2` is very small so by neglecting it we get,

δ  = `-λr_A^2 `

`V(r) = (-e^2)/(4πε_0) = e^((-λδ - λr_A))/((r_A + δ))`

= `(-e^2)/(4πε_0) * e^(-λ(δ + r))/(r_A(1 + δ/r_A))`

= `(-e^2)/(4πε_0r_A) e^(-λr)(1 + δ/r_A)^-1`  ......(∵ r = r_A + δ²)

= `(-e^2e^-(λr))/(4πrε_0r_A) (1 - δ/r_A)`

V(r) = – 27.2 eV remains unchanged

K.E. = `1/2 mv^2 = 1/2 m(h/(mr))^2`

= `1/2 h^2/(mr^2)`  .......[From I, `v = h/(mr)`]

= `(-h^2)/(2m(r_A + δ)^2)`

= `(-h^2)/(2mr_A^2 (1 + δ/r_A)^2`

= `h/(2mr_A^2) (1 - (2δ)/r_A)`

= `h^2/(2mr_A) (1 - (-λr_A^2)/r_A)`

= `h^2/(2mr_A) (1 + 2λr_A)`

= 13.6 eV (1 + 2λr_A)

Total energy = `(-e^2)/(4πε_0r_A) + h^2/(2mr_A^2) (1 + 2λr_A)`

= `[- 27.2 + 13.6 (1 + 2λr_A)] eV`

= `- 27.2 + 13.6 + 27.2  λr_A`

Total E = `- 13.6 + 27.2  λr_A`

Change in energy = – 13.6 + 27.2 λr_A – (– 13.6) = 27.2 λr_A eV