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 (for photon of wavelength λ, energy \({\rm{E}} = \frac{{1240\;{\rm{eV}}}}{{\lambda \left( {{\rm{in\;nm}}} \right)}}\))

An excited He+ ion emits two photons in succession, with wavelengths 1

1.	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 (for photon of wavelength λ, energy \({\rm{E}} = \frac{{1240\;{\rm{eV}}}}{{\lambda \left( {{\rm{in\;nm}}} \right)}}\))
A.	n = 4
B.	n = 5
C.	n = 7
D.	n = 6
Answer» C. n = 7

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