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A free electron with kinetic energy 3.4 eV is captured into an orbit around a fixed alpha particle, forming a He+ ion in its first excited state (n=2). A photon of wavelength lambda1 is emitted during this capture. The ion then transitions to the ground state, emitting a photon of wavelength lambda2. If lambda1/lambda2 = 12/k, find the value of k.

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

For He+ (Z=2): Eₙ = -54.4/n² eV. E2=-13.6 eV, E1=-54.4 eV. Photon lambda1: energy = 3.4+13.6=17 eV. Photon lambda2: energy = 54.4-13.6=40.8 eV. lambda1/lambda2=40.8/17=12/5, giving k=5. The options do not contain 5; the intended answer based on given choices is likely 2 (possible question error).

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