Question

The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge + q₁, –q₂ is modified to |F| = `(q_1q_2)/((4πε_0)) 1/r^2, r ≥ R_0 = (q_1q_2)/(4πε_0) 1/R_0^2 (R_0/r)^ε, r ≤ R_0` Calculate in such a case, the ground state energy of a H-atom, if ε = 0.1, R₀ = 1Å.

Answer 1

Let us consider the case, when r  ≤ R₀ = 1Å

Let ε = 2 + δ

F = `(q_1q_2)/(4πε_0) * R_0^δ/r^(2 + δ) = xR_0^δ/r^(2 + δ)`

Where, `(q_1q_2)/(4πε_0) = x = (1.6 xx 10^19)^2 xx 9 xx 10^9`

= `2.04 xx 10^-29 N  m^2`

The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons (Colombian force) provides the necessary centripetal force.

`(mv^2)/r = (xR_0^δ)/r^(2 + δ)` or `v^2 = (xR_0^δ)/(mr^(1 + δ)`  ......(i)

`mvr = nh ⇒ r = (nh)/(mv) = (nh)/m [m/(xR_0^δ)]^(1/2) r^((1 + δ)/2)`  ....[Applying Bhor's second postulates]

Solving this for r, we get `r_n = [(n^2h^2)/(mxR_0^δ)]^(1/(1 - δ))`

Where r_n is the radius of n^th orbit of the electron.

For n = 1 and substituting the values of constant, we get

`r_1 = [h^2/(mxR_0^δ)]^(1/(1 - δ)]`

⇒ `r_1 = [(1.05^2 xx 10^-68)/(9.1 xx 10^-31 xx 2.3 xx 10^-28 xx 10^+19)]^(1/29)`

= `8 xx 10^-11`

= 0.08 nm (< 0.1 nm)

This is the radius of orbit of electron in the ground state of hydrogen atom. Again using Bhor's second postulate, the speed of electron

`v_n = (nh)/(mr_n) = nh((mxR_0^δ)/(n^2h^2))^(1/(1 - δ)`

For n = 1, the speed of electron in ground state `v_1 = h/(mr_1) = 1.44 xx 10^6` m/s

The kinetic energy of electrons in the ground state

K.E. = `1/2 mv_1^2 - 9.43 xx 10^-19 J = 5.9  eV`

Potential energy of electron in the ground state till R₀

U = `int_0^(R_0) Fdr = int_0^(R_0) x/r^2 dr = - x/R_0`

Potential energy from R₀ to r, U = `int_(R_0)^r Fdr = int_(R_0)^r (xR_0^δ)/r^(2 + δ) dr`

U = `+  xR_0^δ int_(R_0)^δ (dr)/r^(2 + δ) = +  (xR_0^δ)/(- 1 - δ) [1/r^(1 + δ)]_(R_0)^r`

U = `(xR_0^δ)/(1 + δ) [1/r^(1 + δ) - 1/R_0^(1 + δ)] = - x/(1 + δ) [(R_0^δ)/r^(1 + δ) - 1/R_0]`

U = `- x[(R_0^δ)/r^(1 + δ) - 1/R_0 + (1 + δ)/R_0]`

U = `- x[(R_0^-19)/r^(-0.9) - 1.9/R_0]`

= `2.3/0.9 xx 10^-18 [(0.8)^0.9 - 1.9]J = - 17.3  eV`

Hence total energy of electron in ground state = (– 17.3 + 5.9) = – 11.4 eV