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Wave-particle duality — de Broglie and Bohr

(a) An electron is accelerated by V = 1000 V. Calculate the de Broglie wavelength.\n(b) For hydrogen, calculate the radius of the 2nd Bohr orbit and the energy of level n=2.
Given data
V = 1000 Vmₑ = 9.109×10⁻³¹ kge = 1.602×10⁻¹⁹ Ch = 6.626×10⁻³⁴ J·sa₀ = 0.0529 nm (Bohr radius)
Review the theory: Fisica Moderna
Steps
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  1. Calculate the velocity of the electron accelerated by V = 1000 V (in ×10⁷ m/s).
  2. Calculate the de Broglie wavelength λ (in pm = 10⁻¹² m).
  3. Calculate the radius of the 2nd Bohr orbit r₂ (in pm).
  4. Calculate the energy of level n=2 of hydrogen (in eV).
Full worked solution
  1. Calculate the velocity of the electron accelerated by V = 1000 V (in ×10⁷ m/s).
    eV=12mev2⇒v=2eVme=2⋅1.602×10−19⋅10009.109×10−31eV = \tfrac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\dfrac{2eV}{m_e}} = \sqrt{\dfrac{2 \cdot 1.602\times10^{-19} \cdot 1000}{9.109\times10^{-31}}}
    12mev2=eV→v=2eVme=2×1.602×10−19×10009.109×10−31=3.518×1014≈1.876×107 m/s≈6.3%\frac{1}{2}m_e v^2 = eV \rightarrow v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2 \times 1.602\times10^{-19} \times 1000}{9.109\times10^{-31}}} = \sqrt{3.518\times10^{14}} \approx \mathbf{1.876\times10^{7}\,m/s} \approx 6.3\% of the speed of light.
  2. Calculate the de Broglie wavelength λ (in pm = 10⁻¹² m).
    λ=hmev=6.626×10−349.109×10−31⋅1.876×107\lambda = \dfrac{h}{m_e v} = \dfrac{6.626\times10^{-34}}{9.109\times10^{-31} \cdot 1.876\times10^7}
    λ=hmev=6.626×10−349.109×10−31×1.876×107≈3.876×10−11 m=38.76 pm\lambda = \frac{h}{m_e v} = \frac{6.626\times10^{-34}}{9.109\times10^{-31} \times 1.876\times10^{7}} \approx \mathbf{3.876\times10^{-11}\,m} = \mathbf{38.76\,pm} (X-ray scale!). The de Broglie wavelength reveals the wave nature of particles.
  3. Calculate the radius of the 2nd Bohr orbit r₂ (in pm).
    rn=n2a0⇒r2=4⋅52.9 pmr_n = n^2 a_0 \Rightarrow r_2 = 4 \cdot 52.9\,\mathrm{pm}
    rn=n2a0=22×52.9=211.6 pm=0.2116 nmr_n = n^2 a_0 = 2^2 \times 52.9 = \mathbf{211.6\,pm} = \mathbf{0.2116\,nm}. The 2nd Bohr orbit is 4 times larger than the 1st orbit.
  4. Calculate the energy of level n=2 of hydrogen (in eV).
    En=−13.6n2 eV⇒E2=−13.64E_n = -\dfrac{13.6}{n^2}\,\mathrm{eV} \Rightarrow E_2 = -\dfrac{13.6}{4}
    En=−13.6n2 eV→E2=−13.64=−3.4 eVE_n = -\frac{13.6}{n^2}\,\mathrm{eV} \rightarrow E_2 = -\frac{13.6}{4} = \mathbf{-3.4\,eV}. The ionisation energy from n=2n=2 is ∣E2∣=3.4 eV|E_2| = 3.4\,\mathrm{eV}. The transition 2→12\to1 emits 13.6−3.4=10.2 eV13.6 - 3.4 = 10.2\,\mathrm{eV} (Lyman-α\alpha line).
Result:λ_de Broglie = 38.8 pm (X-ray scale) — r₂ = 211.6 pm — E₂ = −3.4 eV.
TheoryFisica ModernaFormula sheetPhysics 2

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