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The energy of a hydrogen-like ion in its nth energy level is represented by En = -Z² × 13.6/n² eV. For the first excited state of He+ with n = 2 and Z = 2, what is the energy?

1. Energy = -4/2² × 13.6 eV = -13.6 eV
2. n(n-1)/2 = 6
3. n² - n - 12 = 0
4. n-4

Correct answer: Energy = -4/2² × 13.6 eV = -13.6 eV

Solution

The energy of the first excited state (n=2) for He+ is calculated using the formula En = -Z² × 13.6/n². Substituting Z=2 and n=2, we get En = -(2² × 13.6)/2² = -13.6 eV.

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