In a hydrogen atom modeled according to Bohr's theory, an electron undergoes a transition … — JAMB UTME

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

Answer: 1.89 eV

Answer

The energy of the emitted photon is

1.89\text{ eV}

According to Bohr's atomic model, when an electron drops from an initial higher energy level

E_i

to a final lower energy level

E_f

, a photon is emitted carrying energy

E = E_i - E_f

. Substituting the given values gives

E = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

Step-by-Step Solution

Identify the initial and final energy states of the electron.

E_i = -1.51\text{ eV}

and

E_f = -3.40\text{ eV}

The electron moves from a higher (less negative) energy state to a lower (more negative) energy state.

Apply Bohr's energy quantization formula for photon emission

E_{\text{photon}} = E_i - E_f

E_{\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.

Key Concept

Bohr's Energy Transition Postulate

Estimated Time:1m 0s

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