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Find the orbit number n of the electron in He⁺ whose de Broglie wavelength equals that of the electron in the 4th orbit of the hydrogen atom.

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

Correct answer: 4

Solution

The de Broglie wavelength in orbit n of a hydrogen-like atom (atomic number Z) is lambda = 2*pi*n*a₀/Z. For hydrogen (Z=1, n=4): lambda_H = 2*pi*4*a₀ = 8*pi*a₀. For He⁺ (Z=2): lambda_He = 2*pi*n*a₀/2 = pi*n*a₀. Setting lambda_He = lambda_H: pi*n*a₀ = 8*pi*a₀, n = 8. Since 8 is not in the options (1,2,3,4), this question appears to have an error in the option set; the correct answer n=8 is not listed.

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