Question

The highest possible energy of a photon in the emission spectrum of hydrogen atom is ______ eV.
[Given: Rydberg constant = 13.61 eV] 
(round off to two decimal places)

Answer 1

Correct Answer: 13.6 - 13.62

Highest Energy of Photon in Hydrogen Emission Spectrum 

The highest possible energy of a photon in the emission spectrum of a hydrogen atom corresponds to the transition where the electron moves from an initial state with principal quantum number \( n_i \) to the final state \( n_f = 1 \). The maximum energy photon is emitted when the electron transitions from the highest possible initial state (\( n_i \rightarrow \infty \)) to \( n_f = 1 \).

The energy difference is given by the Rydberg formula:

\[ E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)^{-1} \]

where \( R_H = 13.61 \, \text{eV} \).

Step 1: When \( n_i \to \infty \)

When \( n_i \to \infty \), \( \frac{1}{n_i^2} \to 0 \), thus the energy difference simplifies to:

\[ E = 13.61 \, \text{eV} \left( \frac{1}{1} - 0 \right) = 13.61 \, \text{eV} \]

Step 2: Final Answer

The energy of the photon is \( 13.61 \, \text{eV} \), which falls within the expected range of \( (13.6, 13.62) \, \text{eV} \).

Conclusion

Therefore, the highest possible energy of a photon in the emission spectrum of a hydrogen atom is 13.61 eV, rounded to two decimal places.