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.

Answer 1

In an H-atom in the ground state, the electron revolves around the point-size proton in a circular orbit of radius r_B (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 < r_B (as r_B = 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 >>  r_B, 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