The de Broglie wavelength of an electron in the $n$th Bohr orbit, in terms of the orbit radius $r_n$, is
A$\lambda = n r_n$
B$\lambda = 2\pi n r_n$
C$\lambda = 2\pi r_n/n$
D$\lambda = r_n/n$
Answer & Solution
Correct answer: C. $\lambda = 2\pi r_n/n$
Bohr's quantisation $2\pi r_n = n\lambda$ is exactly de Broglie's standing-wave condition — an integer number of wavelengths fit around the circumference. Solving: $\lambda = 2\pi r_n/n$.
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