JEE Mains · Physics · STD 12 - 12. atoms
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 \(E = \frac{{1240\,eV}}{{\lambda \,(in\,nm)}}\))
- A \(4\)
- B \(6\)
- C \(5\)
- D \(7\)
Answer & Solution
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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