The de-Broglie's wavelength of an electron in the 4ᵗʰ orbit is ________ π a₀. (a₀= Bohr's radius)

Solution

Chemistry ma’am · 4 steps

Bohr's quantum condition means exactly n waves fit around the orbit, so divide the circumference by n.

1. Waves fit the orbit. Combine Bohr's condition mvr = nh/2π with de Broglie's λ = h/mv. You get 2π r = nλ: a whole number of wavelengths fits around the orbit.

   λ = (2π rₙ)/n

2. Radius of orbit 4. For hydrogen, rₙ = n² a₀. With n = 4, r₄ = 16 a₀.
3. Put in numbers. λ = (2π × 16 a₀)/4 = 8π a₀.
4. So. The de Broglie wavelength in the 4th orbit is 8π a₀, so the answer is 8.

The common slip: 32. Taking the whole circumference 2π r₄ as the wavelength gives 32π a₀, so 32. Four wavelengths fit in the 4th orbit, so divide by 4.

The idea to keep. In Bohr orbits, 2π r = nλ. Since r ∝ n², the de Broglie wavelength is 2π n a₀ for hydrogen, so it grows in step with n.

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