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An excited He^(+) ion emits two photons ...

An excited `He^(+)` ion emits two photons in succession, with wavelength 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 `lamda`, energy `E=(1240eV)/(lamda("in nm"))`

A

n = 4

B

n=5

C

n=6

D

n=7

Text Solution

Verified by Experts

The correct Answer is:
B
`E_("total")=(1240/(108.5)+1240/(30.4))eV = 52.20 " eV"`
`52.20 " eV" = 13.6 xx 4 (1-1/(1^(2)))rArrn^(2)=25 rArrn=5` .
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