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Solved Examples on Atomic Physics Question 1:- A 100 W sodium vapor lamp is placed at the center of a large sphere, which absorbs all the sodium light that falls on it. At what rate are photons delivered to the sphere? The wavelength of sodium light is 590 nm. Solution:- From equation E = hf the energy per photon for sodium light is E = hf = hc/λ = [(6.63×10^{-34 }J.s)(3.00×10^{8 }m/s)]/[590×10^{-9 }m] =3.37×10^{-19 }J Thus, every time a sodium atom emits a photon, the atom loses 3.37×10^{-19 }J, or 2.1 eV, of energy. When the sodium light is absorbed by the sphere, energy is transferred to the sphere in lumps of this same size. To find the rate R at which photons are absorbed by the sphere, we divide E into the rate (the power P) at which the lamp emits energy: R = P/E = [100 W]/[3.37×10^{-19 }J/photon] = 3.0×10^{20} photons/s Thus from the above observation we conclude that, the rate at which the photons are delivered to the sphere would be 3.0×10^{20} photons/s. ____________________________________________________________________________________ Question 2 (JEE Advanced):- X rays of wavelength λ = 22 pm (photon energy = 56 keV) art scattered from a carbon target, and the scattered rays are detected at 85° to the incident beam. (a) What is the Compton shift of the scattered rays? (b) What percentage of the initial x-ray photon energy is transferred to an electron in such scattering? Solution:- From equation Δλ = h/mc [1-cos?] = [(6.63×10^{-34 }J.s) (1-cos 85°)]/[(9.11×10^{-31}kg) (3.00×10^{8}^{ }m/s)] = 2.21×10^{-12} m ≈ 2.2 pm From the above observation we conclude that, the Compton shift of the scattered rays would be 2.2 pm. (b) The fractional energy loss frac is, frac = [E-E'] / E = [hf – hf '] / hf = [c/λ – c/λ'] / [c/λ] = [λ' - λ] / λ' = Δλ / [λ+Δλ] Substitution gives, frac = [2.21 pm]/[22 pm+2.21 pm] = 0.091 or 9.1% Therefore 9.1% of the of the initial x-ray photon energy is transferred to an electron in such scattering. _____________________________________________________________________________________ Question 3:- (a) What is the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines? (b) What is the wavelength corresponding to the series limit for the Lyman series? Solution:- (a) For any series, the transition that produces the least energetic photon is the transition between the home-base level that defines the series and the level immediately above it. For the Lyman series, the transition with the least energetic photon is the transition from the level with n = 2 to that with n = 1. From equation E_{n} = – (me^{4}/8ε_{0}^{2}h^{2}) (1/n^{2}) = – (13.6 eV/n^{2}), for n = 1,2,3,.....the energy difference for this transition is, ΔE = E_{2}-E_{1} = – (13.6 eV) (1/2^{2} – 1/1^{2}) = 10.2 eV The corresponding wavelength is found from equation E = hf, written in the form ΔE = hf = hc/λ, where hf is the energy of the emitted photon. Solving for the wavelength λ gives, λ = hc/ΔE = (6.63×10^{-34 }J.s) (3.00×10^{8}^{ }m/s)/(10.2 eV) (1.60×10^{-19}J/eV) = 1.22×10^{-7} m = 122 nm From the above observation we conclude that, the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines would be 122 nm. Light with this wavelength is in the ultraviolet region of the electromagnetic spectrum. (b) The series limit corresponds to a transition from the level with n = ∞ to that with n = 1, the home base for this series. From equation E_{n} = – (me^{4}/8ε_{0}^{2}h^{2}) (1/n^{2}) = – (13.6 eV/n^{2}), for n = 1,2,3,....., the energy difference for this transition is, ΔE = E_{∞}-E_{1} = – (13.6 eV) [1/∞^{2} – 1/1^{2}] = – (13.6 eV) (0-1) = 13.6 eV The corresponding wavelength is found as in (a) and is λ = hc/ΔE = (6.63×10^{-34 }J.s) (3.00×10^{8}^{ }m/s)/(13.6 eV) (1.60×10^{-19}J/eV) = 9.14×10^{-8} m = 91.4 nm Therefore, the wavelength corresponding to the series limit for the Lyman series would be 91.4 nm. Light with this wavelength is also in the ultraviolet region of the electromagnetic spectrum.
A 100 W sodium vapor lamp is placed at the center of a large sphere, which absorbs all the sodium light that falls on it. At what rate are photons delivered to the sphere? The wavelength of sodium light is 590 nm.
From equation E = hf the energy per photon for sodium light is
E = hf
= hc/λ
= [(6.63×10^{-34 }J.s)(3.00×10^{8 }m/s)]/[590×10^{-9 }m]
=3.37×10^{-19 }J
Thus, every time a sodium atom emits a photon, the atom loses 3.37×10^{-19 }J, or 2.1 eV, of energy. When the sodium light is absorbed by the sphere, energy is transferred to the sphere in lumps of this same size.
To find the rate R at which photons are absorbed by the sphere, we divide E into the rate (the power P) at which the lamp emits energy:
R = P/E
= [100 W]/[3.37×10^{-19 }J/photon]
= 3.0×10^{20} photons/s
Thus from the above observation we conclude that, the rate at which the photons are delivered to the sphere would be 3.0×10^{20} photons/s.
____________________________________________________________________________________
X rays of wavelength λ = 22 pm (photon energy = 56 keV) art scattered from a carbon target, and the scattered rays are detected at 85° to the incident beam.
(a) What is the Compton shift of the scattered rays?
(b) What percentage of the initial x-ray photon energy is transferred to an electron in such scattering?
From equation Δλ = h/mc [1-cos?]
= [(6.63×10^{-34 }J.s) (1-cos 85°)]/[(9.11×10^{-31}kg) (3.00×10^{8}^{ }m/s)]
= 2.21×10^{-12} m
≈ 2.2 pm
From the above observation we conclude that, the Compton shift of the scattered rays would be 2.2 pm.
(b) The fractional energy loss frac is,
frac = [E-E'] / E
= [hf – hf '] / hf
= [c/λ – c/λ'] / [c/λ]
= [λ' - λ] / λ'
= Δλ / [λ+Δλ]
Substitution gives,
frac = [2.21 pm]/[22 pm+2.21 pm]
= 0.091 or 9.1%
Therefore 9.1% of the of the initial x-ray photon energy is transferred to an electron in such scattering.
_____________________________________________________________________________________
(a) What is the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines?
(b) What is the wavelength corresponding to the series limit for the Lyman series?
(a) For any series, the transition that produces the least energetic photon is the transition between the home-base level that defines the series and the level immediately above it. For the Lyman series, the transition with the least energetic photon is the transition from the level with n = 2 to that with n = 1. From equation E_{n} = – (me^{4}/8ε_{0}^{2}h^{2}) (1/n^{2}) = – (13.6 eV/n^{2}), for n = 1,2,3,.....the energy difference for this transition is,
ΔE = E_{2}-E_{1}
= – (13.6 eV) (1/2^{2} – 1/1^{2})
= 10.2 eV
The corresponding wavelength is found from equation E = hf, written in the form
ΔE = hf
= hc/λ,
where hf is the energy of the emitted photon. Solving for the wavelength λ gives,
λ = hc/ΔE
= (6.63×10^{-34 }J.s) (3.00×10^{8}^{ }m/s)/(10.2 eV) (1.60×10^{-19}J/eV)
= 1.22×10^{-7} m
= 122 nm
From the above observation we conclude that, the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines would be 122 nm. Light with this wavelength is in the ultraviolet region of the electromagnetic spectrum.
(b) The series limit corresponds to a transition from the level with n = ∞ to that with n = 1, the home base for this series. From equation E_{n} = – (me^{4}/8ε_{0}^{2}h^{2}) (1/n^{2}) = – (13.6 eV/n^{2}), for n = 1,2,3,....., the energy difference for this transition is,
ΔE = E_{∞}-E_{1}
= – (13.6 eV) [1/∞^{2} – 1/1^{2}]
= – (13.6 eV) (0-1) = 13.6 eV
The corresponding wavelength is found as in (a) and is
= (6.63×10^{-34 }J.s) (3.00×10^{8}^{ }m/s)/(13.6 eV) (1.60×10^{-19}J/eV)
= 9.14×10^{-8} m
= 91.4 nm
Therefore, the wavelength corresponding to the series limit for the Lyman series would be 91.4 nm. Light with this wavelength is also in the ultraviolet region of the electromagnetic spectrum.
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