JEEOlympiad
In a singly ionized helium ion (He+, nuclear charge Z = 2), the…
Question
In a singly ionized helium ion (He+, nuclear charge Z = 2), the electron makes a transition from n = 3 to n = 2. Using 13.6 eV for the hydrogen ground-state binding energy and hc = 1240 eV*nm, the emitted wavelength equals p/q nm in lowest terms. Report p + q.
✓ Verified answer: 2807checked by our engine — not a guess
Step-by-step solution
For He+, E_n = -13.6*Z^2/n^2 with Z=2, so Z^2 = 4.
The n=3 -> n=2 photon energy is 13.6*4*(1/4 - 1/9) = 54.4*(5/36) = 272/36 = 68/9 eV.
The wavelength is lambda = hc/E = 1240/(68/9) = 1240*9/68 = 11160/68 = 2790/17 nm (lowest terms).
Hence p+q = 2790+17 = 2807.
Final answer2807
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