In a hydrogen atom the electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to:

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$\frac{1}{n^{2}}$

$\frac{1}{n^{3}}$

$\frac{1}{n^{4}}$

$\frac{1}{n}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{λ} = R Z^{2} [\frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}}] \\ \Rightarrow f = \frac{c}{λ} = R c Z^{2} [\frac{1}{n^{2}} - \frac{1}{{(n + 1)}^{2}}] \\ \Rightarrow f = R c Z^{2} \frac{(2 n + 1)}{n^{2} {(n + 1)}^{2}} \\ F o r n > > 1 \\ \Rightarrow f \propto \frac{n}{n^{4}} \\ \Rightarrow f \propto \frac{1}{n^{3}} \end{array}$

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