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Difficulty: MediumAtomic Models

In Bohr's model of the hydrogen atom, the energy of an electron in a stationary orbit with principal quantum number nn is given by En=−13.6n2 eVE_n = -\frac{13.6}{n^2}\text{ eV}. What is the frequency of the photon emitted when an electron transitions from the n=4n = 4 energy state to the n=2n = 2 energy state? (h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}, 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. 6.15×1014 Hz6.15 \times 10^{14}\text{ Hz}Answer
  2. B
    3.85×1033 Hz3.85 \times 10^{33}\text{ Hz}
  3. C
    2.05×1014 Hz2.05 \times 10^{14}\text{ Hz}
  4. D
    6.20×1014 Hz6.20 \times 10^{14}\text{ Hz}

Answer

6.15×1014 Hz6.15 \times 10^{14}\text{ Hz}
The energy of the emitted photon corresponds to the difference between the initial energy level (n=4n = 4) and the final energy level (n=2n = 2), giving ΔE=−0.85 eV−(−3.40 eV)=2.55 eV\Delta E = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}. Converting 2.55 eV2.55\text{ eV} to Joules yields 4.08×10−19 J4.08 \times 10^{-19}\text{ J}. Dividing this energy by Planck's constant (6.63×10−34 J⋅s6.63 \times 10^{-34}\text{ J}\cdot\text{s}) produces a photon frequency of 6.15×1014 Hz6.15 \times 10^{14}\text{ Hz}.

Step-by-Step Solution

1
Calculate the energy levels for n=4n = 4 and n=2n = 2
E4=−13.642=−0.85 eVE_4 = -\frac{13.6}{4^2} = -0.85\text{ eV} and E2=−13.622=−3.40 eVE_2 = -\frac{13.6}{2^2} = -3.40\text{ eV}
Electron energy in Bohr's model depends inversely on the square of the principal quantum number.
2
Find the energy of the emitted photon
ΔE=E4−E2=−0.85 eV−(−3.40 eV)=2.55 eV\Delta E = E_4 - E_2 = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}
The energy of the emitted photon equals the energy lost during the downward transition between states.
3
Convert the photon energy from electron-volts to Joules
ΔE=2.55×1.6×10−19 J=4.08×10−19 J\Delta E = 2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}
Energy must be converted to SI units (Joules) before applying Planck's equation.
4
Calculate the photon frequency using f=ΔEhf = \frac{\Delta E}{h}
f=4.08×10−19 J6.63×10−34 J⋅s≈6.15×1014 Hzf = \frac{4.08 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.15 \times 10^{14}\text{ Hz}
Photon energy and frequency are related by the Planck relation E=hfE = h f.

Key Concept

Bohr Model Energy Transitions and Photon Frequency
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Topics

Atomic Models

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