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Zorluk: OrtaAtomic Models

In a hydrogen atom modeled according to Bohr's theory, an electron undergoes a transition from an excited state with an energy of −1.51 eV-1.51\text{ eV} to a lower energy state of −3.40 eV-3.40\text{ eV}. Calculate the energy of the emitted photon in electron-volts (eV\text{eV}).

Cevap: 1.89 eV

Cevap

The energy of the emitted photon is 1.89 eV1.89\text{ eV}.
According to Bohr's atomic model, when an electron drops from an initial higher energy level EiE_i to a final lower energy level EfE_f, a photon is emitted carrying energy E=Ei−EfE = E_i - E_f. Substituting the given values gives E=−1.51 eV−(−3.40 eV)=1.89 eVE = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}.

Adım Adım Çözüm

1
Identify the initial and final energy states of the electron.
Ei=−1.51 eVE_i = -1.51\text{ eV} and Ef=−3.40 eVE_f = -3.40\text{ eV}.
The electron moves from a higher (less negative) energy state to a lower (more negative) energy state.
2
Apply Bohr's energy quantization formula for photon emission Ephoton=Ei−EfE_{\text{photon}} = E_i - E_f.
Ephoton=−1.51−(−3.40)=1.89 eVE_{\text{photon}} = -1.51 - (-3.40) = 1.89\text{ eV}.
By energy conservation, the energy lost by the transitioning electron equals the energy of the emitted photon.

Anahtar Kavram

Bohr's Energy Transition Postulate
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Atomic Models

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