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What is the de Broglie wavelength associated with an electron in the n = 4 energy level of a hydrogen atom, expressed in terms of the de Broglie wavelength of the electron in the ground state (n = 1)?

1. two times the de Broglie wavelength in the ground state
2. four times the de Broglie wavelength in the ground state
3. half of the de Broglie wavelength in the ground state
4. one-fourth of the de Broglie wavelength in the ground state

Correct answer: four times the de Broglie wavelength in the ground state

Solution

Bohr's quantization condition states that the circumference of the nth orbit equals n de Broglie wavelengths: 2*pi*rₙ = n*lambdaₙ. Since rₙ = n² * a0 (where a0 is Bohr radius), we get lambdaₙ = 2*pi*n²*a0 / n = 2*pi*n*a0. Therefore lambdaₙ is proportional to n. For n = 4: lambda₄ = 4 * lambda₁. The de Broglie wavelength in n = 4 is four times that in n = 1.

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