Advertisements
Advertisements
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.
Advertisements
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
APPEARS IN
संबंधित प्रश्न
Using Bohr's postulates of the atomic model, derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohr's radius.
Using Bohr's postulates, derive the expression for the total energy of the electron in the stationary states of the hydrogen atom ?
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.
When a photon stimulates the emission of another photon, the two photons have
(a) same energy
(b) same direction
(c) same phase
(d) same wavelength
In which of the following systems will the wavelength corresponding to n = 2 to n = 1 be minimum?
According to Bohr, 'Angular momentum of an orbiting electron is quantized'. What is meant by this statement?
If l3 and l2 represent angular momenta of an orbiting electron in III and II Bohr orbits respectively, then l3: l2 is :
Draw energy level diagram for a hydrogen atom, showing the first four energy levels corresponding to n=1, 2, 3 and 4. Show transitions responsible for:
(i) Absorption spectrum of Lyman series.
(ii) The emission spectrum of the Balmer series.
According to Bohr's theory, an electron can move only in those orbits for which its angular momentum is integral multiple of ____________.
Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). Also show that for large values of n, this frequency equals to classical frequency of revolution of an electron.
The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because ______.
The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?
The ground state energy of hydrogen atoms is -13.6 eV. The photon emitted during the transition of electron from n = 3 to n = 1 unknown work function. The photoelectrons are emitted from the material with a maximum kinetic energy of 9 eV. Calculate the threshold wavelength of the material used.
Given below are two statements:
Statements I: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement II: According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increase with a decrease in principal quantum number.
In light of the above statements, choose the most appropriate answer from the options given below:
A 100 eV electron collides with a stationary helium ion (He+) in its ground state and exits to a higher level. After the collision, He+ ions emit two photons in succession with wavelengths 1085 Å and 304 Å. The energy of the electron after the collision will be ______ eV.
Given h = 6.63 × 10-34 Js.
The energy of an electron in the first Bohr orbit of the H-atom is −13.6 eV. The energy value of an electron in the excited state of Li2+ is ______.
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)
In hydrogen atom, transition from the state n = 6 to n = 1 results in ultraviolet radiation. Infrared radiation will be obtained in the transition ______.
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.
