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प्रश्न
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 mp, force would be modified to |F| = `e^2/((4πε_0)r^2) [1/r^2 + λ/r]`, exp (– λr) where λ = mpc/h and h = `h/(2π)`. Estimate the change in the ground state energy of a H-atom if mp were 10-6 times the mass of an electron.
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उत्तर
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` >> rA (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)`
∵ λA >>> rB and r = ra + δ
Taking rA + r + λr2 .....[∵ r = (ra + δ]
rA = (rA + δ) + λ(rA + δ) ....(Put r = rA + 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 rA << 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 = rA + 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 + 2λrA)
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 λrA – (– 13.6) = 27.2 λrA eV
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