Wavelength of de Broglie wave of electron in n-th Bohr orbit:
Aλ = 2π r_n / n (n complete wavelengths fit around orbit)
Bλ = h × n
Cλ = r_n
Dλ = h/n
Answer & 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.
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