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The transition in the hydrogen atom from the second orbit to the first orbit emits a photon of a certain wavelength. The same wavelength is emitted when an electron in the He+ ion transitions from the nth orbit to the second orbit. Find the value of n.

1. n = 2
2. n = 4
3. n = 3
4. n = 6

Correct answer: n = 4

Solution

For hydrogen (Z=1), transition from n=2 to n=1: 1/lambda = R*(1)²*(1/1² - 1/2²) = R*(1 - 1/4) = 3R/4. For He+ (Z=2), transition from n to n=2: 1/lambda = R*(2)²*(1/2² - 1/n²) = 4R*(1/4 - 1/n²). Setting them equal: 3R/4 = 4R*(1/4 - 1/n²). Dividing by R: 3/4 = 1 - 4/n². Therefore 4/n² = 1 - 3/4 = 1/4. So n² = 16, giving n = 4.

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