The de Broglie wavelength of an electron in the 4th Bohr orbit is :

Solution

Chemistry ma’am · 3 steps

Fit whole waves around the orbit: 2π rₙ = nλ, then put in rₙ = n² a₀.

1. Standing wave. In Bohr's model the electron wave must close on itself, so a whole number of wavelengths fits around the orbit: 2π rₙ = nλ.
2. Orbit radius. For the hydrogen atom, rₙ = n² a₀, where a₀ is the Bohr radius (0.529 Å). For n = 4, r₄ = 16a₀.

   λ = (2π rₙ)/n = (2π n² a₀)/n = 2π n a₀

3. So. With n = 4, λ = 2π × 4a₀ = 8π a₀, which is option (c).

The idea to keep. In the nth Bohr orbit of hydrogen, λ = 2π n a₀. The wavelength grows in step with n, because the radius grows as n² while you fit n waves around it.

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