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9702 · 22.4

Energy levels in atoms and line spectra — practice questions

Practice and worked examples for 9702 Energy levels in atoms and line spectra. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

An electron in a hydrogen atom de-excites from an energy level of -1.51 eV to a ground state level of -13.6 eV. Calculate the frequency of the emitted photon. (Use h=6.63×10−34 Jsh = 6.63 \times 10^{-34} \text{ Js} and 1 eV=1.60×10−19 J1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}).

Show solution outline

  1. Find the energy difference (in eV): ΔE=Ephoton=Einitial−Efinal=−1.51 eV−(−13.6 eV)=12.09 eV \Delta E = E_{photon} = E_{initial} - E_{final} = -1.51 \text{ eV} - (-13.6 \text{ eV}) = 12.09 \text{ eV}

  2. Convert the energy difference to Joules: ΔE=12.09 eV×(1.60×10−19 J/eV)=1.9344×10−18 J \Delta E = 12.09 \text{ eV} \times (1.60 \times 10^{-19} \text{ J/eV}) = 1.9344 \times 10^{-18} \text{ J}

  3. Use the formula E=hfE = hf to find the frequency (ff): f=ΔEh=1.9344×10−18 J6.63×10−34 Jsf = \frac{\Delta E}{h} = \frac{1.9344 \times 10^{-18} \text{ J}}{6.63 \times 10^{-34} \text{ Js}}

  4. Calculate the frequency: f≈2.92×1015 Hz f \approx 2.92 \times 10^{15} \text{ Hz}

    The emitted photon has a frequency of approximately 2.92×1015 Hz2.92 \times 10^{15} \text{ Hz}.

Worked example 2

Light of wavelength 486 nm is observed in the emission spectrum of hydrogen. This corresponds to a transition to the n=2 energy level, which has an energy of -3.40 eV. Calculate the energy of the initial, higher energy level in eV. (Use h=6.63×10−34 Jsh = 6.63 \times 10^{-34} \text{ Js}, c=3.00×108 m s−1c = 3.00 \times 10^{8} \text{ m s}^{-1}, and 1 eV=1.60×10−19 J1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}).

Show solution outline

  1. Calculate the energy of the emitted photon in Joules (J): The relationship between photon energy, wavelength (λλ), and the speed of light (cc) is E=hc/λE = hc/λ. First, convert the wavelength to metres: 486 nm=486×10−9 m486 \text{ nm} = 486 \times 10^{-9} \text{ m}. Ephoton=(6.63×10−34 Js)×(3.00×108 m s−1)486×10−9 m=4.0926×10−19 JE_{photon} = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3.00 \times 10^{8} \text{ m s}^{-1})}{486 \times 10^{-9} \text{ m}} = 4.0926 \times 10^{-19} \text{ J}

  2. Convert the photon energy to electronvolts (eV): Ephoton(eV)=4.0926×10−19 J1.60×10−19 J/eV=2.558 eVE_{photon} (\text{eV}) = \frac{4.0926 \times 10^{-19} \text{ J}}{1.60 \times 10^{-19} \text{ J/eV}} = 2.558 \text{ eV}

  3. Determine the initial energy level (EinitialE_{initial}): The energy of the emitted photon is the difference between the initial and final energy levels: Ephoton=Einitial−EfinalE_{photon} = E_{initial} - E_{final}. We know Ephoton=2.558 eVE_{photon} = 2.558 \text{ eV} and Efinal=−3.40 eVE_{final} = -3.40 \text{ eV}. 2.558 eV=Einitial−(−3.40 eV)2.558 \text{ eV} = E_{initial} - (-3.40 \text{ eV}) 2.558 eV=Einitial+3.40 eV2.558 \text{ eV} = E_{initial} + 3.40 \text{ eV}

  4. Solve for EinitialE_{initial}: Einitial=2.558 eV−3.40 eV=−0.842 eVE_{initial} = 2.558 \text{ eV} - 3.40 \text{ eV} = -0.842 \text{ eV}

    The energy of the initial, higher energy level is approximately -0.84 eV.

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