alleryThe de Broglie wavelength $\lambda$ of an electron in the $n$th Bohr orbit of hydrogen, in terms of the orbit radius $r_n$, is
A$\lambda = 2\pi n r_n$
B$\lambda = \tfrac{2\pi r_n}{n}$
C$\lambda = n r_n$
D$\lambda = \tfrac{r_n}{n}$
Answer & Solution
Correct answer: B. $\lambda = \tfrac{2\pi r_n}{n}$
Bohr's stable-orbit condition is exactly the de Broglie standing-wave condition: the orbit circumference equals an integer number of wavelengths, $2\pi r_n = n\lambda$.
Solving for $\lambda$ gives $\tfrac{2\pi r_n}{n}$.
This is the bridge between Bohr's ad-hoc quantisation and wave mechanics.
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