Quantum Physics and Light

Quantum - Chapter 1 - Light Quanta

The Electromagnetic Spectrum

  • Wavelength is measured in meters.
  • Frequency is measured in Hertz (Hz).
  • The electromagnetic spectrum includes:
    • Radio waves: Wavelength 10710^7 m, Frequency 10410^4 Hz
    • Microwaves
    • Infrared
    • Visible light (human sight)
    • Ultraviolet
    • X-rays
    • Gamma rays
  • As wavelength decreases, frequency increases.

Blackbody Radiation

  • Blackbody Radiation Curves (BBRC):

    • As temperature (T) increases, the peak of the BBRC shifts to greater intensities and shorter wavelengths.
    • Electromagnetic waves (EMWs) emit from a hot body due to the oscillation of electric charges in the molecules of the material.

  • Wave Model for Black Body (Rayleigh and Jeans Rule):

    • Wave theory predicts the variation of spectral radiancy (intensity) with wavelength (λ) at a given temperature.
    • S=2πckTλ4S = \frac{2\pi c kT}{\lambda^4}
      • where k is Boltzmann constant
    • The wave theory agrees with experiment at very long wavelengths but disagrees at shorter wavelengths.
    • Intensity of electromagnetic radiation tends to zero when wavelength increases (infrared region, IR), but approaches infinity when wavelength decreases (Ultraviolet region, UV).

  • Blackbody Definition:

    • Absorbs all thermal radiation incident on its surface and emits radiation of wavelengths (λ\lambda) that depends only on its temperature (T).
    • An ideal blackbody can be formed by making a cavity within a body.
    • Its thermal emissivity (e) = 1.

Wien’s Law

  • The relationship between the temperature T of a black body and the wavelength λmax\lambda_{max} at which the intensity is maximum is inversely proportional.
  • λmax×T=2898×10−6m⋅K\lambda_{max} \times T = 2898 \times 10^{-6} m \cdot K
  • T∝1λmaxT \propto \frac{1}{\lambda_{max}}
Example Problems using Wien's Law
  • Problem 1:
    • At what temperature is cavity radiation most visible to the human eye if the eye is most sensitive to yellow-green light, wavelength = 550 nm?
    • Solution:
      • λmax=550nm\lambda_{max} = 550 nm
      • Using Wien's Law: λmax×T=2898×10−6m⋅K\lambda_{max} \times T = 2898 \times 10^{-6} m \cdot K
      • T=2898×10−6550×10−9=5269.09KT = \frac{2898 \times 10^{-6}}{550 \times 10^{-9}} = 5269.09 K
  • Problem 2:
    • The effective surface temperature of the sun is 5800 K. At what wavelength would you expect the sun to radiate most strongly? In what region of the spectrum is this? Why, then, does the sun appear yellow?
    • Solution:
      • T = 5800 K
      • Using Wien's Law: λmax×T=2898×10−6m⋅K\lambda_{max} \times T = 2898 \times 10^{-6} m \cdot K
      • λmax=2898×10−65800=0.4997×10−6m=499.7nm\lambda_{max} = \frac{2898 \times 10^{-6}}{5800} = 0.4997 \times 10^{-6} m = 499.7 nm
      • This is in the visible region (yellow color).
      • The sun appears yellow because it radiates most strongly at the wavelength of yellow color.
  • Problem 3:
    • Find the peak wavelength of the radiation emitted by each of the following:
      • a) Human body when the skin temperature is 35°C
      • b) The tungsten filament, of a light bulb which operates at 2000 °K
    • Solution:
      • a) Human body:
        • T=273+35=308KT = 273 + 35 = 308 K
        • λmax×308=2898×10−6m⋅K\lambda_{max} \times 308 = 2898 \times 10^{-6} m \cdot K
        • λmax=2898×10−6308=9.4×10−6m=9.4μm\lambda_{max} = \frac{2898 \times 10^{-6}}{308} = 9.4 \times 10^{-6} m = 9.4 \mu m
        • This radiation is in the infrared region (IR), which is the basis for night vision instruments.
      • b) Tungsten filament:
        • T=2000KT = 2000 K
        • λmax×2000=2898×10−6m⋅K\lambda_{max} \times 2000 = 2898 \times 10^{-6} m \cdot K
        • λmax=2898×10−62000=1.4×10−6m=1.4μm\lambda_{max} = \frac{2898 \times 10^{-6}}{2000} = 1.4 \times 10^{-6} m = 1.4 \mu m

Stefan-Boltzmann Law

  • Relates the thermal power of a black body to its temperature.
  • P=σAe(T<em>24−T</em>14)P = \sigma A e (T<em>2^4 - T</em>1^4)
    • P is the Power radiated (W)
    • σ\sigma is the Stefan's Constant = 5.67x10−8(Wm−2K−4)5.67 x 10^{-8} (W m^{-2} K^{-4})
    • A is the Surface area of the body (m²)
    • T2T_2 is the Temperature of the body (K)
    • T1T_1 is the room Temperature (K)
    • e is the thermal emissivity of the body, ranging from 0 to 1 depending on the nature of the blackbody surface.
Example Problem using Stefan-Boltzmann Law
  • An oven with an inside temperature 227 °C is in a room having a temperature 27 °C. There is a small opening of area 5 cm² in one side of the oven. How much net power is transferred from the oven to the room? Consider both oven and room as cavities.
  • Solution:
    • T2=227+273=500KT_2 = 227 + 273 = 500 K
    • T1=27+273=300KT_1 = 27 + 273 = 300 K
    • A=5cm2=5×10−4m2A = 5 cm^2 = 5 \times 10^{-4} m^2
    • e = 1
    • σ=5.67×10−8(Wm−2K−4)\sigma = 5.67 \times 10^{-8} (W m^{-2} K^{-4})
    • P=(5.67×10−8)(5×10−4)(1)((500)4−(300)4)=1.54WP = (5.67 \times 10^{-8})(5 \times 10^{-4})(1)((500)^4 - (300)^4) = 1.54 W
Additional Example
  • If the stars behave like black bodies, for the sun λ<em>max=510nm\lambda<em>{max} = 510 nm whereas for the north-star λ</em>max=350nm\lambda</em>{max} = 350 nm. Find:
    • (a) The surface temperature of these stars.
    • (b) Using Stefan's Boltzmann law and the result obtained above, determine the power radiated from 1 cm² of the star surface. Given σ=5.67×10−8W/m2⋅k4\sigma = 5.67 \times 10^{-8} W/m^2 \cdot k^4
  • Solution:
    • (a)
      • For the sun:
        • λmax⋅T=2898μm⋅K\lambda_{max} \cdot T = 2898 \mu m \cdot K
        • 510×10−9⋅T=2898×10−6m⋅K510 \times 10^{-9} \cdot T = 2898 \times 10^{-6} m \cdot K
        • T=5700KT = 5700 K
      • For the north-star:
        • 350×10−9⋅T=2898×10−6m⋅K350 \times 10^{-9} \cdot T = 2898 \times 10^{-6} m \cdot K
        • T=8300KT = 8300 K
    • (b)
      • For the sun:
        • Power=eσAT4Power = e \sigma A T^4
        • Power=1×1×10−4×5.67×10−8×57004=5.9×103WPower = 1 \times 1 \times 10^{-4} \times 5.67 \times 10^{-8} \times 5700^4 = 5.9 \times 10^3 W
      • For the north-star:
        • Power=1×1×10−4×5.67×10−8×83004=2.71×104WPower = 1 \times 1 \times 10^{-4} \times 5.67 \times 10^{-8} \times 8300^4 = 2.71 \times 10^4 W

Photon Model for Black Body - Plank's Quantum Hypothesis

  • Max Planck derived a formula for the spectral radiancy that fitted the experimental data perfectly at all wavelengths and for all temperatures.
  • Planck's Quantum Hypothesis:
    1. Blackbody radiation is due to oscillating molecules at its surface.
    2. Energy of molecules (E<em>nE<em>n) have only discrete values given by: E</em>n=nhfE</em>n = n h f
    3. The molecules emit or absorb energy in discrete packets (photons) by jumping between quantum states.
  • I(λ,T)=2πhc2λ51ehcλkT−1I(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda kT}} - 1}
  • En=nhfE_n = n h f
  • ΔE=hCλ=E<em>upper−E</em>lower\Delta E = \frac{hC}{\lambda} = E<em>{upper} - E</em>{lower}
Example: Planck's Law tends to Rayleigh-Jeans law for long wavelength
  • Show that Planck's law tends to Rayleigh-Jeans law for long wavelengths (200 angstrom).
  • I(λ,T)=2πhc2λ5×1ehcλkBT−1I(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \times \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1}
  • For long λ\lambda, the exponential term in the denominator can be approximated by the first two terms in the series:
    • ex=1+x+x22!+…e^x = 1 + x + \frac{x^2}{2!} + …
    • ex≈1+xe^x \approx 1 + x
  • ehcλkT=1+hcλkTe^{\frac{hc}{\lambda kT}} = 1 + \frac{hc}{\lambda kT}
  • I(λ,T)=2πhc2λ5×11+hcλkT−1I(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \times \frac{1}{1 + \frac{hc}{\lambda kT} - 1}
  • I(λ,T)=2πhc2λ5×λkThc=2πckTλ4I(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \times \frac{\lambda kT}{hc} = \frac{2 \pi c kT}{\lambda^4}

Light Quanta (Photon)

  • Light consists of small particles called light quanta (Photons), each with a specified energy and momentum.
  • The energy of a single photon (E) is:
    • E=hf=hCλE = hf = \frac{hC}{\lambda}
    • where h is the Plank constant, h=6.63×10−34J⋅s=4.14×10−15eV⋅sh = 6.63 \times 10^{-34} J \cdot s = 4.14 \times 10^{-15} eV \cdot s
    • 1.0eV=1.6×10−19J1.0 eV = 1.6 \times 10^{-19} J
  • The photon has momentum (P):
    • P=hλ=EcP = \frac{h}{\lambda} = \frac{E}{c}
  • E=mc2E = mc^2
  • p=mcp = mc
  • p=hλp = \frac{h}{\lambda}
  • E=p⋅cE = p \cdot c
  • p=Ecp = \frac{E}{c}
Example Problems: Photon Energy and Momentum

  • Problem 1:

    • An atomic oscillator emits radiation of 700 nm wavelength. How much energy is associated with a photon of light of this wavelength?
    • Solution:
      • E=hcλ=6.626×10−34J⋅s×3×108m/s700×10−9m=2.84×10−19JE = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} J \cdot s \times 3 \times 10^8 m/s}{700 \times 10^{-9} m} = 2.84 \times 10^{-19} J

  • Problem 2:

    • Yellow light from a sodium vapor lamp has an effective Wavelength of 589 nm. What is the energy of the corresponding photons?
    • Solution:
      • λ=589nm\lambda = 589 nm
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • E=hCλ=4.14×10−15eV⋅s×3×108m/s589×10−9m=2.11eVE = \frac{hC}{\lambda} = \frac{4.14 \times 10^{-15} eV \cdot s \times 3 \times 10^8 m/s}{589 \times 10^{-9} m} = 2.11 eV

  • Problem 3:

    • During radioactive decay, a certain nucleus emits a gamma ray whose photon energy is 1.35 MeV.
      • (a) To what wavelength does this photon correspond?
      • (b) what is the momentum of this photon?
    • Solution:
      • E=1.35MeVE = 1.35 MeV
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • (a) λ=hCE=(4.14×10−15)(3×108)1.35×106=9.2×10−13m\lambda = \frac{hC}{E} = \frac{(4.14 \times 10^{-15})(3 \times 10^8)}{1.35 \times 10^6} = 9.2 \times 10^{-13} m
      • (b) P=EC=1.35×1063×108=4.5×10−3eV⋅s/mP = \frac{E}{C} = \frac{1.35 \times 10^6}{3 \times 10^8} = 4.5 \times 10^{-3} eV \cdot s/m

  • Problem 4:

    • Show that the energy E of a photon (in eV) is related to its wavelength λ (in nm) by

      • E(eV)=1240λ(nm)E(eV) = \frac{1240}{\lambda(nm)}

    • The orange-colored light from a highway sodium lamp has a wavelength of 589 nm. How much energy is possessed by an individual photon from such a lamp?

    • Solution:

      • E=hf=hcλ=(4.14×10−15eV⋅s)(3×108×109nm/s)λE = hf = \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15} eV \cdot s)(3 \times 10^8 \times 10^9 nm/s)}{\lambda}
      • E=hcλ=1240λ(nm)E = \frac{hc}{\lambda} = \frac{1240}{\lambda (nm)}
      • λ=589nm\lambda = 589 nm
      • E=1240589=2.1eVE = \frac{1240}{589} = 2.1 eV

  • Problem 5:

    • A particular x-ray photon has a wavelength of 35 pm. calculate the photon’s (a) energy, (b) frequency, and (c) momentum.
    • Solution:
      • λ=35pm=0.035nm\lambda = 35 pm = 0.035 nm
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • (a) E=hcλ=12400.035=35428.6eVE = \frac{hc}{\lambda} = \frac{1240}{0.035} = 35428.6 eV
      • (b) f=cλ=3×10835×10−12=8.57×1018Hzf = \frac{c}{\lambda} = \frac{3 \times 10^8}{35 \times 10^{-12}} = 8.57 \times 10^{18} Hz
      • (c) p=hλ=4.14×10−1535×10−12=1.183×10−4eV⋅s/mp = \frac{h}{\lambda} = \frac{4.14 \times 10^{-15}}{35 \times 10^{-12}} = 1.183 \times 10^{-4} eV \cdot s/m

Photoelectric Effect

  • When light is incident on a metallic surface, the surface can emit electrons (Photoelectrons).
  • Light energy converts to electric energy.
  • E_{photon} = hf > \Phi
    • whereΦ\Phi is the Work function
  • Ephoton=Φ+K.E.E_{photon} = \Phi + K.E.
Experimental Setup & Observations
  • When a beam of light falls on a metal plate in an evacuated tube:
    • Photoelectrons are ejected from the metal surface if Eincident≥ΦE_{incident} \geq \Phi, where Φ\Phi is the work function of the metal.
    • Photoelectrons are collected by the collector (C).
    • A suitable potential difference (V) between the emitter (E) and (C) sweeps up these photoelectrons.
    • If the potential of the circuit is reversed and increased, no photoelectrons move to the collector (Stopping Potential).
Stopping Potential (Vs)
  • It is the negative voltage at which no photocurrent or photoelectrons move.
  • At stopping potential (Vs):
    • K<em>max=12m</em>eν2=eVsK<em>{max} = \frac{1}{2} m</em>e \nu^2 = eV_s
    • me=9.11×10−31kgm_e = 9.11 \times 10^{-31} kg is the mass of the electron
    • ν\nu is the speed of the photoelectron
    • e=1.6×10−19Ce = 1.6 \times 10^{-19} C is the charge of the electron
    • KmaxK_{max} = Maximum kinetic energy
Work Function (Φ)
  • It is the minimum energy needed to remove an electron from the surface of a metal.
  • Cut off frequency (fof_o)
    • Emission of electrons from the surface can only occur if hf≥ϕhf \geq \phi
    • Cut-off frequency is the minimum frequency of the incident light needed to emit photoelectrons.
    • Φ=hf<em>o=hcλ</em>o\Phi = hf<em>o = \frac{hc}{\lambda</em>o}
    • If hf≤Φhf \leq \Phi then no emission occurs.
    • Φ=hf<em>o=hcλ</em>o\Phi = hf<em>o = \frac{hc}{\lambda</em>o}
Photoelectric Current & Voltage Graph
  • VsV_s is independent of the intensity of the incident light.
  • The photocurrent increases with the intensity of incident light.
  • No current (photocurrent = 0) flows at −V≤−Vs-V \leq -V_s
  • The kinetic energy K<em>max=0K<em>{max} = 0 of photoelectrons at cutoff frequency (f</em>of</em>o) is zero.
  • kinetic energy of photoelectrons increases with frequency of light: Kmax∝E=hfK_{max} \propto E = hf
Einstein’s Explanation
  • Each photon gives its energy to an electron in the metal.
  • E=hf=hcλ=1240λ(nm)eVE = hf = \frac{hc}{\lambda} = \frac{1240}{\lambda (nm)} eV
  • This energy equals:
    • hf=Φ+Kmaxhf = \Phi + K_{max}
    • Where KmaxK_{max} is the maximum kinetic energy of the photoelectron
    • Φ\Phi is called the work function of the metal
  • Maximum Kinetic energy of ejected electron:
    • K<em>max=12mv2=eV</em>sK<em>{max} = \frac{1}{2} mv^2 = eV</em>s
  • Energy of incident photon
    • E=hf=hcλ=1240λ(nm)eVE = hf = \frac{h c}{\lambda} = \frac{1240}{\lambda (nm)} eV
  • Work function of the metal
    • Φ=hf<em>O=hcλ</em>O=1240λc(nm)eV\Phi = hf<em>O = \frac{h c}{\lambda</em>O} = \frac{1240}{\lambda_c (nm)} eV
Important notes
  • K<em>max=eV</em>sK<em>{max} = eV</em>s
    • e=1.6×10−19Ce = 1.6 \times 10^{-19} C is the charge of the electron when K<em>max=eV</em>sK<em>{max} = eV</em>s measure by Joel
    • e = 1 is charge of electron when K<em>max=eV</em>sK<em>{max}=eV</em>s measure by electron volt (eV)
  • It must be using the unit of KmaxK_{max} in Joel to calculate the speed of photoelectron:
    • K<em>max=eV</em>s=1.6×10−19CVsVolt=…JoelK<em>{max} = eV</em>s = 1.6 \times 10^{-19}C V_s Volt = …Joel
    • =(E−Φ)/1.6×10−19C=12mev2= (E - \Phi)/1.6 \times 10^{-19}C = \frac{1}{2} m_e v^2
Example Problems: Photoelectric Effect
  • Problem 8:
    • The energy needed to remove an electron from metallic sodium is 2.28 eV.
      • (i) Does sodium show a photoelectric effect for red light, with λ = 680 nm?
      • (ii) what is the cutoff wavelength for photoelectric emission from sodium?
    • Solution:
      • (i) E = \frac{hc}{\lambda} = \frac{1240}{\lambda (nm)} = \frac{1240}{680} = 1.82 eV < \Phi
        • So, sodium doesn’t show the photoelectric for red light
      • (ii) Φ=hf<em>o=hcλ</em>o\Phi = hf<em>o = \frac{hc}{\lambda</em>o}
        • λo=hcΦ=12402.28=543.86nm\lambda_o = \frac{hc}{\Phi} = \frac{1240}{2.28} = 543.86 nm
  • Problem 9:
    • Find the maximum kinetic energy of photoelectrons if the work function of the material is 2.3 eV and the frequency of the radiation is 3 × 10¹⁵ Hz.
    • Solution:
      • f=3×1015Hzf = 3 \times 10^{15} Hz
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • Φ=2.3eV\Phi = 2.3 eV
      • E<em>photon=Φ+K</em>maxE<em>{photon} = \Phi + K</em>{max}
      • Kmax=hf−Φ=4.14×10−15⋅3×1015−2.3=10.108eVK_{max} = hf - \Phi = 4.14 \times 10^{-15} \cdot 3 \times 10^{15} - 2.3 = 10.108 eV
  • Problem 11:
    • The work function of tungsten is 4.5 eV. Calculate the speed of the fastest of the photoelectrons emitted when photons of energy 5.8 eV are incident on a sheet of tungsten.
    • Solution:
      • E=5.8eVE = 5.8 eV
      • Φ=4.5eV\Phi = 4.5 eV
      • m=9.11×10−31kgm = 9.11 \times 10^{-31} kg
      • E=Φ+12mev2E = \Phi + \frac{1}{2} m_e v^2
      • v=2(E−Φ)me=2(5.8−4.5)×1.6×10−19J9.11×10−31kg=6.76×105m/sv = \sqrt{\frac{2(E - \Phi)}{m_e}} = \sqrt{\frac{2(5.8 - 4.5) \times 1.6 \times 10^{-19} J}{9.11 \times 10^{-31} kg}} = 6.76 \times 10^5 m/s
  • Problem 12:
    • Light of wavelength 200 nm falls on an aluminum surface. In Aluminum 4.2 eV are required to remove an electron.
      • (a) What is the kinetic energy of the fastest and (b) the slowest emitted photoelectrons?
      • (c) What is the stopping potential?
      • (d) What is the cutoff wavelength for aluminum?
    • Solution:
      • λ=200nm\lambda = 200 nm
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • Φ=4.2eV\Phi = 4.2 eV
      • (a) E=Φ+KmaxE = \Phi + K_{max}
        • Kmax=hcλ−Φ=1240200−4.2=2.0eVK_{max} = \frac{hc}{\lambda} - \Phi = \frac{1240}{200} - 4.2 = 2.0 eV
      • (b) K.E.slowest=ZeroK.E._{slowest} = Zero
      • (c) K<em>max=eV</em>sK<em>{max} = eV</em>s
        • 2(eV)=e×Vs2 (eV) = e \times V_s
        • Vs=2voltV_s = 2 volt
      • (d) Φ=hcλo\Phi = \frac{hc}{\lambda_o}
        • λo=hcΦ=12404.2=295.24nm\lambda_o = \frac{hc}{\Phi} = \frac{1240}{4.2} = 295.24 nm
  • Problem 13:
    • (a) If the work function for a metal is 1.8 eV, what would be the stopping potential for light having a wavelength of 400 nm?
      • (b) what would be the maximum speed of the emitted photoelectrons at the metal’s surface?
    • Solution:
      • λ=400nm\lambda = 400 nm
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • Φ=1.8eV\Phi = 1.8 eV
      • m=9.11×10−31kgm = 9.11 \times 10^{-31} kg
      • E=Φ+KmaxE = \Phi + K_{max}
        • 1240400=1.8+Kmax\frac{1240}{400} = 1.8 + K_{max}
        • K<em>max=eV</em>s=1.3eVK<em>{max} = e V</em>s = 1.3 eV
        • Vs=1.3voltV_s = 1.3 volt
      • K<em>max=hcλ−Φ=12m</em>ev2K<em>{max} = \frac{hc}{\lambda} - \Phi = \frac{1}{2} m</em>e v^2
        • v=2K<em>maxm</em>e=2×1.3×1.6×10−19J9.11×10−31kg=6.76×105m/sv = \sqrt{\frac{2 K<em>{max}}{m</em>e}} = \sqrt{\frac{2 \times 1.3 \times 1.6 \times 10^{-19} J}{9.11 \times 10^{-31} kg}} = 6.76 \times 10^5 m/s
  • Problem 15:
    • The stopping potential for photoelectrons emitted from a metal surface illuminated by light of wavelength 491 nm is 0.71 V. When the incident wavelength is changed to a new value, the stopping potential is found to be 1.43 V.
      • (a) what is his new wavelength?
      • (b) what is the work function for the surface?
    • Solution:
      • λ1=491nm\lambda_1 = 491 nm
      • h=4.14×10−15eV⋅sh = 4.14 \times 10^{-15} eV \cdot s
      • Vs1=0.71VV_{s1} = 0.71 V
      • Vs2=1.43VV_{s2} = 1.43 V
      • hcλ<em>1=Φ+eV</em>s1\frac{hc}{\lambda<em>1} = \Phi + eV</em>{s1}
      • Φ=hcλ<em>1−eV</em>s1=1240481−1×0.71=1.87eV\Phi = \frac{hc}{\lambda<em>1} - eV</em>{s1} = \frac{1240}{481} - 1 \times 0.71 = 1.87 eV
      • hcλ<em>2=Φ+eV</em>s2\frac{hc}{\lambda<em>2} = \Phi + eV</em>{s2}
      • λ<em>2=hcΦ+eV</em>s2=12401.87+1×1.43=376.06nm\lambda<em>2 = \frac{hc}{\Phi + eV</em>{s2}} = \frac{1240}{1.87 + 1 \times 1.43} = 376.06 nm

Compton Effect

  • The shift in wavelengths ([Delta]λ = λsc - λ) was measured experimentally.
  • The scattered photon has a longer wavelength (λsc) than the incident wavelength.
  • When an X-ray photon (λ) collides with a target of graphite, the electrons are ejected from the atom or molecule with a scattered photon of wavelength (λsc).
Experimental Results
  • The shift in wavelengths ([Delta]λ = λsc - λ) increases with the angle of detection.
  • Wavelength of the scattered photon is found to be different from that of the incident photon and it depends on the detection angle φ:
    • The incident photon causes the electron to oscillate with the same frequency.
    • The electron is a charged particle, so its oscillation radiates an electromagnetic wave with the same frequency.
  • Explanation of wave model:
    • There is no shift in frequency or wavelength [Delta]λ = λsc - λ = 0, which is opposite of experimental observations.
Compton Explanation (Photon model)

  • The X-ray photon gives part of its energy to the electron in the form of kinetic energy.

  • The rest part of the X-ray photon energy is converted to a scattered photon of energy.

  • So, the shift in wavelength is ([Delta]λ).

  • The Compton equation is:

    • Δλ=λ<em>s−λ=hmC(1−cosφ)=λ</em>c(1−cosθ)\Delta \lambda = \lambda<em>s - \lambda = \frac{h}{mC}(1 - cos \varphi) = \lambda</em>c (1 - cos \theta)
    • Δλ\Delta \lambda is called the Compton shift
    • m is the mass of the electron
    • λc is called the Compton wavelength = 2.43 pm

  • X-ray incident photon has energy E=hcλE = \frac{h c}{\lambda} and momentum P=hλP = \frac{h}{\lambda}colliding with free electron E=mc2E = mc^2

  • \lambda_{sc} > \lambda

Important angles
  • When (φ = 0), Grazing collision, the incident photon pass without any scattering, Δλ=0\Delta \lambda = 0
  • When (φ = 180°), Head-on collision, the incident photon reverses its direction, Δλ=λ<em>c(1−Cos180)=2λ</em>c\Delta \lambda = \lambda<em>c (1 - Cos 180) = 2 \lambda</em>c
Derivation of Compton equation:
  • Apply energy and linear momentum conservation rule between the photon and electron.
    • i) Energy conservation equation: hCλ=hCλs+mc2(γ−1)\frac{hC}{\lambda} = \frac{hC}{\lambda_s} + mc^2 (\gamma-1)
    • ii) The momentum equation:
      • X – components: hλ=hλscosφ+γmvcosθ\frac{h}{\lambda} = \frac{h}{\lambda_s} cos \varphi + \gamma mv cos \theta
      • Y – components: hλssinφ=γmvsinθ\frac{h}{\lambda_s} sin\varphi = \gamma mv sin \theta
  • From equations 1, 2, & 3
  • Δλ=λs−λ=hmc(1−cosφ)\Delta \lambda = \lambda_s - \lambda = \frac{h}{mc} (1 -cos\varphi)
Compton Effect Problems
  • Example 6: X-rays of wavelength 22 pm (photon energy = 56 keV) are scattered from a carbon target, radiation being viewed at 85° to the incident beam.
    • (a) What is the Compton shift?
    • (b) what percentage of this initial energy does an incident x ray photon loose?
    • Solution:
      • (a) Given:
        • λ=22pm\lambda = 22 pm
        • E=56keVE = 56 keV
        • λc=2.43pm\lambda_c = 2.43 pm
        • ϕ=85°\phi = 85°
        • Required: Δλ,E−EsE=?\Delta \lambda, \frac{E - Es}{E} = ?
      • Δλ=λc(1−cosϕ)=2.43(1−cos85°)=2.21pm\Delta \lambda = \lambda_c (1 - cos \phi) = 2.43 (1 - cos 85°) = 2.21 pm
      • E−EsE=hf−hfshf=Cλ−Cλ<em>sCλ=λ×λ</em>s−λλλs=λs−λλs\frac{E-Es}{E} = \frac{hf - hfs}{hf} = \frac{\frac{C}{\lambda} - \frac{C}{\lambda<em>s}}{\frac{C}{\lambda}} = \frac{\lambda \times \lambda</em>s - \lambda}{\lambda{\lambda_s}}= \frac{\lambda s - \lambda}{\lambda s}
      • ∴E−EsE=λs−λλs=Δλλ+Δλ=2.2122+2.21=0.091=9.1\therefore \frac{E - Es}{E} = \frac{\lambda s - \lambda}{\lambda s} = \frac{\Delta \lambda}{\lambda + \Delta \lambda} = \frac{2.21}{22 + 2.21} = 0.091 = 9.1 %
  • Problem 17: Photons of wavelength 2.4 pm are incident on a target containing free electrons.
    • (a) Find the wavelength of a photon that is scattered at 30° from the incident direction (b) do the same for a scattering angle of 120°.
    • Solution:
      • Given: $$"lambda