Bohr–de Broglie relation: The number of standing de Broglie waves fitting an electron’s circular orbit equals the principal quantum number <em>n</em>. If the maximum magnetic quantum number is +3 (so l = 3), determine the number of waves for the lowest allowed orbit with that l. Choose the correct option.

Given data

• Maximum magnetic quantum number m_l = +3 ⇒ orbital angular momentum quantum number l = 3.

Concept/ApproachFor hydrogenic orbits, n ≥ l + 1. The number of standing waves around the orbit equals n (since 2πr = nλ).

Step-by-stepl = 3 ⇒ the smallest allowed principal quantum number is n = l + 1 = 4.Therefore, number of waves = n = 4.

Final Answer4