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Difficulty: EasyEnergy Levels and Atomic Spectra

An electron in a hydrogen atom undergoes a transition from an energy state of −1.51 eV-1.51\text{ eV} to a lower energy state of −3.40 eV-3.40\text{ eV}. What is the energy of the emitted photon in electron-volts (eV\text{eV})?

Answer: 1.89 eV

Answer

The energy of the emitted photon is 1.89 eV1.89\text{ eV}.
The energy of an emitted photon during an atomic transition is given by ΔE=Einitial−Efinal\Delta E = E_{\text{initial}} - E_{\text{final}}. Substituting the given levels yields ΔE=−1.51 eV−(−3.40 eV)=1.89 eV\Delta E = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}.

Step-by-Step Solution

1
Identify the initial and final energy states.
Ei=−1.51 eVE_i = -1.51\text{ eV} and Ef=−3.40 eVE_f = -3.40\text{ eV}
The energy of the photon corresponds to the difference between these two levels.
2
Apply the energy conservation formula for atomic emission.
Ephoton=Ei−EfE_{\text{photon}} = E_i - E_f
When an electron drops to a lower energy level, a photon carrying the lost energy is released.
3
Substitute the values and evaluate the difference.
Ephoton=−1.51−(−3.40)=1.89 eVE_{\text{photon}} = -1.51 - (-3.40) = 1.89\text{ eV}
Subtracting the negative lower energy value yields a positive photon energy.

Key Concept

Photon energy from atomic level transitions
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Topics

Energy Levels and Atomic Spectra

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