An excited \(He^+\) ion emits two photons in succession, with wavelengths \(108.5\, nm\) and \(30.4\, nm\), in making a transition to ground state. The quantum number \(n\), corresponding to its initial excited state is \(n=........\) (for photon of wavelength \(\lambda \), energy \(E = \frac{{1240\,eV}}{{\lambda \,(in\,nm)}}\))

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Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\(\frac{1}{\lambda}=R\left(\frac{1}{m^{2}}-\frac{1}{n^{2}}\right) z^{2}\)

\(\frac{1}{1085}=R\left(\frac{1}{m^{2}}-\frac{1}{n^{2}}\right) 2^{2}\)

\(\therefore m=2\)

\(\therefore n=5\)

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