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Question

Find the quantum number n corresponding to the state of He+ ion, if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4 nm (take RH=1.09×105 cm1)

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Solution

Given, D2=30.4×107cm
D1=108.5×107cm
Let, excited state of He+ be n2. It comes from n2 to n1 and then n1 to 1 to emit two succesive photon
1D2=RH.Z2[1121n21]130.4×107=1.09×105×4[1121n21]n1=2Now for D1:n1=2 and n2:?1D1=RHZ2[1221n22]1108.5×107=1.09×105×4×[1221n22]n2=5
Thus, excited state for He is 5th orbit.

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