If $n_{1} + n_{2} = 4$ and $n_{2}^{2} - n_{1}^{2} = 8$ , then calculate maximum value of wavelength emitted in transition form $n_{2} \to n_{1}$ for ${Li}^{2 +}$ in nm [Given $R_{H} = 10^{7} m^{- 1}]$ .

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answer is 80.

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Detailed Solution

We have given that

$\begin{array}{l} n_{2} = 3, n_{1} = 1 \\ \frac{1}{λ} = R_{H} \times 3^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}] \\ \Rightarrow λ = \frac{4}{5 R_{H}} = 8 \times 10^{- 8} m = 80 nm \end{array}$

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