In a sample of hydrogen atom in ground state electrons make transition from ground state to a particular excited state where path length $(2 π r)$ is five times the de-broglie’s wavelength. When electrons make back transition to the ground state producing all possible photons. If photon having 2^nd highest energy of this sample is exactly used to excite the electron in a particular excited state of $L i^{2 +}$ ion then the final excited state of $L i^{2 +}$ ion is_______

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answer is 12.

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Detailed Solution

Since electron goes the state where the path length is 5 times de-broglie wavelength.
$\Rightarrow 2 π r = 5 λ$ also $\frac{2 π r}{n} = λ \Rightarrow n = 5$
Hence electron goes to the 5th state.
Question Image
2nd highest energy line will be 4  1
If this photon is used for Li+2 then, 13.6 $(1 - \frac{1}{4^{2}}) = 13.6 \times Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$n_{1} n_{2} \Rightarrow 3 \to 12$
Hence final excited state = 12.