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Question: Answered & Verified by Expert
An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is



(for a photon of wavelength λ, energy E=1240 eVλ(in nm))
PhysicsAtomic PhysicsJEE MainJEE Main 2019 (12 Apr Shift 1)
Options:
  • A n=6
  • B n=5
  • C n=7
  • D n=4
Solution:
1570 Upvotes Verified Answer
The correct answer is: n=5
E1=1240λ1=1240108.511.43eV
E2=1240λ2=124030.440.79eV
ETotal=E1+E1=52.22eV
52.22=13.6221-1n2
52.22=54.41-1n2
0.96=1-1n2
1n2=0.04
n2=10.04=1004=25
n=5

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