Atomic Structure: Practice Problem & Solution
Energy E of a hydrogen atom with principal quantum number n is given by $E = \frac{-13.6}{n^2} eV$. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately: (2004)
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:
The energy of the emitted photon is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) eV$. This gives $\Delta E = 13.6 \times \frac{5}{36} = 1.88 eV \approx 1.9 eV$.
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