Question

The number of de Broglie waves that fit into the n^th Bohr orbit is:

• A. n²
• B. n
• C. n - 2
• D. n³

Hint:

de Broglie's standing wave condition sets orbit circumference equal to n wavelengths.

Answer 1

The Correct Option is B

Step 1: Bohr proposed that an electron can only move in certain fixed orbits where its angular momentum is quantised: \(mvr_n = \dfrac{nh}{2\pi}\), where n is a positive integer, the principal quantum number.

Step 2: Louis de Broglie later explained why only these particular orbits are allowed. He proposed that a moving electron behaves like a wave with wavelength \(\lambda = \dfrac{h}{mv}\).

Step 3: For the electron wave to exist as a stable standing wave around the orbit, so that it does not cancel itself out after going around, the total circumference of the orbit, \(2\pi r_n\), must equal exactly a whole number of wavelengths: \(2\pi r_n = n\lambda\).

Step 4: Substituting \(\lambda = \dfrac{h}{mv}\) into this condition gives \(2\pi r_n = \dfrac{nh}{mv}\), which rearranges to \(mvr_n = \dfrac{nh}{2\pi}\). This is exactly Bohr's angular momentum rule, so the two ideas match perfectly.

Step 5: The number of complete de Broglie waves that fit around the nth orbit is \(\dfrac{2\pi r_n}{\lambda} = n\). For n = 1, one wave fits, for n = 2, two waves fit, and so on. It is a direct one to one match with the orbit number, not n squared or n cubed.

Answer: The number of de Broglie waves is n, option B.