Picture the electron's wave wrapped around the circular Bohr orbit like a loop of string tied end to end. For the wave to survive without destroying itself by interference, the loop must close up smoothly, meaning the starting point and the point after going all the way around must be in phase.
- This is only possible if the circumference of the orbit is an exact whole number multiple of the electron's wavelength: circumference $= n \times \lambda$, with $n = 1, 2, 3, ...$
- If the circumference were, say, $2.5\lambda$, the wave would be out of step with itself after one loop and would destructively interfere and cancel out. Such an orbit is not allowed, which is exactly why Bohr's orbits are quantised.
- Using de Broglie's relation $\lambda = h/(mv)$, the allowed orbit condition $2\pi r_n = n\lambda$ becomes $mvr_n = nh/(2\pi)$, identical to Bohr's original angular momentum rule that he had proposed without explanation.
- Directly from $2\pi r_n = n\lambda$, the number of full wavelengths, that is full waves, that fit exactly around the orbit is $n$ itself, nothing squared or cubed.
So the orbit number n and the wave count are the same number: the nth orbit carries exactly n de Broglie waves, confirming option B.