An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10⁻¹¹ m while that for the electron in third excited state is 8.48 × 10⁻¹⁰ m. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

Solution

Physics sir · 4 steps

Fit whole wavelengths around each orbit: λ = 2π r/n, and the radii tell you n.

1. Find n. In Bohr's model rₙ ∝ n². Here (8.48 × 10⁻¹⁰)/(5.3 × 10⁻¹¹) = 16 = n², so n = 4. That fits, because the third excited state is n = 4.
2. Bohr condition. A stable orbit holds a whole number of de Broglie wavelengths, so 2π rₙ = nλₙ.

   λₙ = (2π rₙ)/n ∝ n²/n = n

3. Compare the two. λ₄/λ₁ = r₄/r₁ × 1/4 = 16/4 = 4. The excited-state wavelength is 4 times the ground-state one. Strictly, ground to excited is 1:4. The options give only this factor of 4.
4. So. The two de Broglie wavelengths differ by a factor of 4, which is option (a).

The idea to keep. In Bohr orbits λ = 2π r/n, and since r ∝ n², λ ∝ n. The radius ratio gives you n², the wavelength ratio gives you n.

Answer from the NTA final answer key. Solution by PrepWiser.