Photon Emission

Question

Photon frequency for transition n=3 to n=2 in H: ν =? (E_n = -13.6/n² eV; h = 4.14×10^{-15} eV·s)

RankGuru · Accepted Answer

1. **Energy Transition**: When an electron jumps from $n=3$ to $n=2$, it emits a photon with energy $\Delta E = E_3 - E_2$.
2. **Calculate Energies**:
   - $E_3 = -13.6 / 3^2 = -13.6 / 9 \approx -1.51 	ext{ eV}$
   - $E_2 = -13.6 / 2^2 = -13.6 / 4 = -3.40 	ext{ eV}$
3. **Energy Difference**:
   - $\Delta E = (-1.51) - (-3.40) = 1.89 	ext{ eV}$ (or more precisely $13.6 	imes [1/4 - 1/9] = 13.6 	imes 5/36 \approx 1.889 	ext{ eV}$).
4. **Frequency Formula**: $E = h
u \implies 
u = E / h$.
5. **Substitution**:
   - $
u = 1.889 / (4.14 	imes 10^{-15}) \approx 0.456 	imes 10^{15} 	ext{ Hz} = 4.56 	imes 10^{14} 	ext{ Hz}$.
6. **Result**: Rounded to two decimal places, the frequency is **$4.57 	imes 10^{14}$ Hz**.

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