The de Broglie wavelength $\lambda$ of an electron in the $n$th Bohr orbit of hydrogen, in terms of the orbit radius $r_n$, is
- $\lambda = 2\pi n r_n$
- $\lambda = \tfrac{2\pi r_n}{n}$
- $\lambda = n r_n$
- $\lambda = \tfrac{r_n}{n}$
Show answer and 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.