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A free electron with kinetic energy 3.4 eV is captured into an orbit around a stationary nucleus, forming a He+ ion. The ion subsequently de-excites to the ground state in two steps, emitting photons of wavelengths lambda1 and lambda2 respectively. If lambda1/lambda2 = 12/k, find the value of k.

1. k = 1
2. k = 3
3. k = 7
4. k = 9

Correct answer: k = 3

Solution

The electron (KE = 3.4 eV) is captured into n=4 of He+ (since E₄ = -3.4 eV). The two-step de-excitation is n=4 to n=2 (delta_E1 = 10.2 eV, giving lambda1) and n=2 to n=1 (delta_E2 = 40.8 eV, giving lambda2). Since lambda is proportional to 1/delta_E: lambda1/lambda2 = delta_E2/delta_E1 = 40.8/10.2 = 4 = 12/3, so k = 3.

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