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In an excited hydrogen atom, when the electron jumps from nth energy level to (n-1)th energy level, wavelength of emitted radiation is $λ$ , when the electron jumps from (n-1)th energy level to (n-2)th energy level, wavelength of emitted radiation is $K λ$ . Then what will be the wavelength of emitted radiation when the electron jumps from nth energy level to (n-2)th energy level?

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$\frac{Kλ}{(K + 1)}$

$\frac{K^{2} λ}{(K+1)}$

$\frac{λ}{(K+1)}$

$(\frac{K-1}{K+1}) λ$

answer is A.

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Detailed Solution

$\frac{hc}{λ} = 13 .6 [\frac{1}{{(n - 1)}^{2}} - \frac{1}{n^{2}}]$

$\frac{hc}{Kλ} = 13 .6 [\frac{1}{{(n - 1)}^{2}} - \frac{1}{{(n - 2)}^{2}}]$

$∴$ and $\frac{hc}{λ^{1}} = 13 .6 [\frac{1}{{(n - 2)}^{2}} - \frac{1}{n^{2}}]$

$∴ \frac{hc}{λ^{1}} = 13 .6 [\frac{1}{{(n - 2)}^{2}} - \frac{1}{{(n - 1)}^{2}}] + 13 .6 [\frac{1}{{(n - 1)}^{2}} - \frac{1}{n^{2}}]$

$\Rightarrow \frac{hc}{λ^{1}} = \frac{hc}{Kλ} + \frac{hc}{λ}$

$\Rightarrow λ^{1} = \frac{Kλ}{(K + 1)}$

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