Atomic Structure: Practice Problem & Solution
In a hypothetical situation, all the atoms in a hydrogen sample are excited to same state. During de-excitation, photon with lowest energy was found to have \(0.66\text{ eV}\). The photon with the highest energy will have energy equal to
Solution Explained:
To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:
For hydrogen atom, \(E_n - E_{n-1} = 0.66\text{ eV}\) corresponds to \(n = 5\) to \(n = 4\) transition (since \(E_5 - E_4 = -0.85 - (-1.51) = 0.66\text{ eV}\)). The highest energy photon is emitted for transition from \(n = 5\) to \(n = 1\), which is \(E_5 - E_1 = -0.85 - (-13.6) = 12.75\text{ eV}\).
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