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Q.

The shell number n corresponding to the excited state of $H e^{+}$ ion if on transition to
the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm is

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answer is 5.

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Detailed Solution

$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \Rightarrow \frac{1}{108.5 \times 10^{- 7}} + \frac{1}{30.4 \times 10^{- 7}} = 1.09 \times 10^{7} \times 2^{2} \times (\frac{1}{l^{2}} - \frac{1}{n^{2}}) ∴ n = 5$

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