Advertisements
Advertisements
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 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.
Advertisements
Solution
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
APPEARS IN
RELATED QUESTIONS
Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is 2.14 × 10-10 m and 1.09 × 106 m/s respectively. [π= 3.142]
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.
The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?
Using Bohr’s postulates for hydrogen atom, show that the total energy (E) of the electron in the stationary states tan be expressed as the sum of kinetic energy (K) and potential energy (U), where K = −2U. Hence deduce the expression for the total energy in the nth energy level of hydrogen atom.
The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?
According to Bohr, 'Angular momentum of an orbiting electron is quantized'. What is meant by this statement?
Mention demerits of Bohr’s Atomic model.
The spectral line obtained when an electron jumps from n = 5 to n = 2 level in hydrogen atom belongs to the ____________ series.
The energy associated with the first orbit of He+ is ____________ J.
According to Bohr’s theory, the angular momentum of an electron in 5th orbit is ______.
The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because ______.
Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom ______.
- because of energy conservation.
- without simultaneously releasing energy in the from of radiation.
- because of momentum conservation.
- because of angular momentum conservation.
Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about ______.
Use Bohr's postulate to prove that the radius of nth orbit in a hydrogen atom is proportional to n2.
A hydrogen atom in is ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of ______.
(Given, Planck's constant = 6.6 × 10-34 Js)
Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength λ. If R is the Rydberg constant then the principal quantum number n of the excited state is ______.
Specify the transition of an electron in the wavelength of the line in the Bohr model of the hydrogen atom which gives rise to the spectral line of the highest wavelength ______.
The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4861 Å. Calculate the wavelength of the first line of the same series.
How much is the angular momentum of an electron when it is orbiting in the second Bohr orbit of hydrogen atom?
Energy and radius of first Bohr orbit of He+ and Li2+ are:
[Given RH = −2.18 × 10−18 J, a0 = 52.9 pm]
