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Difficulty: MediumEnergy Levels and Atomic Spectra

An electron in an excited state of an atom moves from an energy level of −2.80 eV-2.80\text{ eV} to a lower energy state of −7.60 eV-7.60\text{ eV}. What is the energy of the emitted photon in Joules? (1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. 7.68×10−19 J7.68 \times 10^{-19}\text{ J}Answer
  2. B
    4.80 J4.80\text{ J}
  3. C
    1.66×10−18 J1.66 \times 10^{-18}\text{ J}
  4. D
    1.22×10−18 J1.22 \times 10^{-18}\text{ J}

Answer

7.68×10−19 J7.68 \times 10^{-19}\text{ J}
The emitted photon's energy is given by the difference between the initial higher state and the final lower state: ΔE=E2−E1=−2.80 eV−(−7.60 eV)=4.80 eV\Delta E = E_2 - E_1 = -2.80\text{ eV} - (-7.60\text{ eV}) = 4.80\text{ eV}. Converting to Joules gives 4.80×1.6×10−19 J=7.68×10−19 J4.80 \times 1.6 \times 10^{-19}\text{ J} = 7.68 \times 10^{-19}\text{ J}.

Step-by-Step Solution

1
Calculate the energy difference between the initial and final states in electron-volts (eV)
ΔE=Einitial−Efinal=−2.80 eV−(−7.60 eV)=4.80 eV\Delta E = E_{\text{initial}} - E_{\text{final}} = -2.80\text{ eV} - (-7.60\text{ eV}) = 4.80\text{ eV}
The energy of an emitted photon during a downward atomic transition equals the difference in energy between the two states.
2
Convert the photon energy from electron-volts (eV) to Joules (J)
E=4.80×1.6×10−19 J=7.68×10−19 JE = 4.80 \times 1.6 \times 10^{-19}\text{ J} = 7.68 \times 10^{-19}\text{ J}
SI units require energy to be expressed in Joules, where 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J}.

Key Concept

Energy level transitions and photon emission
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Topics

Energy Levels and Atomic Spectra

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