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Difficulty: EasyEnergy Levels and Atomic Spectra

An electron in a hydrogen atom transitions from an excited energy state of −3.4 eV-3.4\text{ eV} to the ground state of −13.6 eV-13.6\text{ eV}. What is the energy of the photon emitted during this transition?

  1. 10.2 eV10.2\text{ eV}Answer
  2. B
    17.0 eV17.0\text{ eV}
  3. C
    −10.2 eV-10.2\text{ eV}
  4. D
    3.4 eV3.4\text{ eV}

Answer

10.2 eV10.2\text{ eV}
When an electron drops to a lower energy state, the energy of the emitted photon equals the difference between the initial and final energy levels: Ephoton=Einitial−Efinal=−3.4 eV−(−13.6 eV)=10.2 eVE_{photon} = E_{initial} - E_{final} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV}.

Step-by-Step Solution

1
Identify the initial and final energy states.
Initial state Ei=−3.4 eVE_i = -3.4\text{ eV} and final ground state Ef=−13.6 eVE_f = -13.6\text{ eV}.
The electron drops from the higher energy state to the lower state.
2
Calculate the photon energy using the transition formula Ephoton=Ei−EfE_{photon} = E_i - E_f.
Ephoton=−3.4 eV−(−13.6 eV)=10.2 eVE_{photon} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV}.
By energy conservation, the energy of the emitted photon must equal the energy loss of the electron.

Key Concept

Photon Emission in Atomic Energy Level Transitions
Estimated Time:45s
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Topics

Energy Levels and Atomic Spectra

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