Electrons in a sample of H-atoms make transition from state $n = x$ to some lower excited state. The emission spectrum from the sample is found to contain only the line belonging to a particular series. If one of the photons has an energy of $0.6375$ eV, find the value of x.
[ Take $0.6375 eV = \frac{3}{4} \times 0.85 eV$ ]

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answer is 8.

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Detailed Solution

n = x to n = ?

$\begin{array}{l} 0.6375 = \frac{3}{4} \times 0.85 = \frac{3}{4} \times \frac{13.6}{16} = \frac{13.6}{4^{2}} [\frac{1}{1^{2}} - \frac{1}{2^{2}}] \\ = 13.6 [\frac{1}{4^{2}} - \frac{1}{8^{2}}] \\ \Rightarrow x = 8 \end{array}$

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