Wavelength of de Broglie wave of electron in n-th Bohr orbit:
- λ = 2π r_n / n (n complete wavelengths fit around orbit)
- λ = h × n
- λ = r_n
- λ = h/n
Show answer and solution
Correct answer: A. λ = 2π r_n / n (n complete wavelengths fit around orbit)
**Bohr:** m v r = nh/(2π) → p = nh/(2π r). λ = h/p = 2π r/n.
So circumference 2π r = n λ — orbit fits exactly n de Broglie wavelengths.
Origin of standing-wave/Bohr quantization.