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Photoelectric effect

UV light with λ = 180 nm strikes a caesium surface with φ = 2.0 eV.\nCalculate: (a) the photon energy in eV, (b) the maximum KE of the emitted electron, (c) the maximum speed of electrons, (d) the threshold frequency.
Given data
λ = 180×10⁻⁹ mφ = 2.0 eV = 3.204×10⁻¹⁹ Jh = 6.626×10⁻³⁴ J·sc = 3×10⁸ m/smₑ = 9.109×10⁻³¹ kg1 eV = 1.602×10⁻¹⁹ J
Review the theory: Fisica Moderna
Steps
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  1. Calculate the photon energy in eV.
  2. Calculate the maximum KE of the emitted electron (in eV).
  3. Calculate the maximum speed of the emitted electrons (in ×10⁶ m/s).
  4. Calculate the threshold frequency ν₀ (in ×10¹⁴ Hz) below which no electrons are emitted.
Full worked solution
  1. Calculate the photon energy in eV.
    Efotone=hcλ=6.626×10−34⋅3×108180×10−9E_{fotone} = \dfrac{hc}{\lambda} = \dfrac{6.626\times10^{-34} \cdot 3\times10^8}{180\times10^{-9}}
    E=hcλ=6.626×10−34×3×108180×10−9=1.988×10−251.8×10−7=1.104×10−18 JE = \frac{hc}{\lambda} = \frac{6.626\times10^{-34} \times 3\times10^8}{180\times10^{-9}} = \frac{1.988\times10^{-25}}{1.8\times10^{-7}} = 1.104\times10^{-18}\,\mathrm{J}. In eV: 1.104×10−181.602×10−19≈6.90 eV\frac{1.104\times10^{-18}}{1.602\times10^{-19}} \approx \mathbf{6.90\,eV}.
  2. Calculate the maximum KE of the emitted electron (in eV).
    KEmax=Efotone−ϕ=6.90−2.0KE_{max} = E_{fotone} - \phi = 6.90 - 2.0
    KEmax⁡=E−ϕ=6.90−2.0=4.90 eVKE_{\max} = E - \phi = 6.90 - 2.0 = \mathbf{4.90\,eV}. The excess photon energy beyond the work function becomes the kinetic energy of the emitted electron (Einstein's photoelectric equation).
  3. Calculate the maximum speed of the emitted electrons (in ×10⁶ m/s).
    vmax=2⋅KEmaxme=2⋅4.90⋅1.602×10−199.109×10−31v_{max} = \sqrt{\dfrac{2 \cdot KE_{max}}{m_e}} = \sqrt{\dfrac{2 \cdot 4.90 \cdot 1.602\times10^{-19}}{9.109\times10^{-31}}}
    KE=4.90×1.602×10−19=7.850×10−19 JKE = 4.90 \times 1.602\times10^{-19} = 7.850\times10^{-19}\,\mathrm{J}. v=2KEme=2×7.85×10−199.109×10−31=1.724×1012≈1.313×106 m/sv = \sqrt{\frac{2KE}{m_e}} = \sqrt{\frac{2 \times 7.85\times10^{-19}}{9.109\times10^{-31}}} = \sqrt{1.724\times10^{12}} \approx \mathbf{1.313\times10^{6}\,m/s}.
  4. Calculate the threshold frequency ν₀ (in ×10¹⁴ Hz) below which no electrons are emitted.
    hν0=ϕ⇒ν0=ϕh=2.0⋅1.602×10−196.626×10−34h\nu_0 = \phi \Rightarrow \nu_0 = \dfrac{\phi}{h} = \dfrac{2.0 \cdot 1.602\times10^{-19}}{6.626\times10^{-34}}
    ν0=ϕh=2.0×1.602×10−196.626×10−34=3.204×10−196.626×10−34≈4.835×1014 Hz\nu_0 = \frac{\phi}{h} = \frac{2.0 \times 1.602\times10^{-19}}{6.626\times10^{-34}} = \frac{3.204\times10^{-19}}{6.626\times10^{-34}} \approx \mathbf{4.835\times10^{14}\,Hz} (near UV). Below this frequency no electrons are emitted.
Result:E_photon = 6.90 eV — KE_max = 4.90 eV — v_max = 1.31×10⁶ m/s — ν₀ = 4.84×10¹⁴ Hz.
TheoryFisica ModernaFormula sheetPhysics 2

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