If the radius of the first orbit of hydrogen atom is α₀, then de Broglie's wavelength of electron in 3ʳᵈ orbit is :

Solution

Chemistry ma’am · 4 steps

In a Bohr orbit, n de Broglie waves fit exactly around the circumference.

1. Standing wave condition. 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. Radius of third orbit. Bohr radii grow as n²: rₙ = n²α₀. For n = 3, r₃ = 9α₀.
3. Put in numbers. λ = (2π r₃)/3 = (2π × 9α₀)/3 = 6πα₀.
4. So. The de Broglie wavelength in the third orbit is 6πα₀, option (c).

The idea to keep. For any Bohr orbit, use 2π r = nλ with r = n²α₀. That gives λ = 2π nα₀, so the wavelength grows in direct proportion to n.

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