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Question
If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å, and (ii) R = 10 Å.
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Solution
In an H-atom in the ground state, the electron revolves around the point-size proton in a circular orbit of radius rB (Bohr’s radius).
As `mvr_B = h` and `(mv^2)/r_B = (-1 xx e xx e)/(4πε_0r_B^2)`
`m/r_B ((-h^2)/(m^2r_B^2)) = e^2/(4πε_0r_B^2)`
`r_B = (4πε_0 h^2)/e^2 h^2/m` = 0.53 Å
K.E. = `1/2 mv^2 = (m/2)(h/(mr_B))^2`
= `h^2/(2mr_B^2)` = 13.6 eV
PE of the electron and proton,
U = `1/(4πε_0) (e(- e))/r_B = - e^2/(4πε_0r_B)` = – 27.2 eV
The total energy of the electron, i.e.,
E = K + U = +13.6 eV – 27.2 eV = – 13.6 eV
(i) When R = 0.1 Å: R < rB (as rB = 0.51 Å) and the ground state energy is the same as obtained earlier for point-size proto 13.6 eV
(ii) When R = 10 A: R >> rB, the electron moves inside the proton (assumed to be a sphere of radius R) with new Bohr’s radius r'B
Clearly, `r_B^' = (4πε_0 h^2)/(m(e)(e^'))` .....[Replacing e2 by (e) (e’) where e’ is the charge on the sphere of radius r'B]
Since `e^' = [e/(((4pi)/3)R^3)] [((4pi)/3) r_B^('3)] = r_B^('3)/R^3`
`r_B^' = (4πε_0 h^2)/(m_e (er_B^('3) R^3)) = ((4πε_0 h^2)/(me^2))(R^3/r_B^('3))`
or `r_B^('4) = ((4πε_0 h^2)/(me^2)) R^3`
= (0.51 Å) (10 Å)3 = (510 Å4)
∴ `r_B^'` = 4.8 Å, which is less than R(= 10 Å)
KE of the electron,
`K^' = 1/2 mv^('2) = (m/2) (h^2/(m^2r_B^('2))) = h/(2mr_B^('2))`
= `(h^2/(2mr_B^2)) (r_B/r_B^')` = 13.6 eV `((0.51 Å)/(4.8 Å))^2` = 0.16 eV
Potential at a point inside the charged proton i.e.,
`V = (k_e e)/R (3 - (r_B^('2))/R^2) = k_e e ((3R^2 - r_B^('2))/R^3)`
Potential energy of electron and proton, μ = – eV
= `- e(k_ee) [(3R^2 - r_B^('2))/R^3]`
= `-(e^2/(4πε_0 r^B)) [(r_B(3R^2 - r_B^('2)))/R^3]`
= `- (27.2 eV) [((0.51 Å)(300 Å - 23.03 Å))/((1000 Å))]`
= – 3.83 eV
Total energy of the electron, E = K + U = 0.16 eV – 3.83 eV = – 3.67 eV
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