An electron is present in an orbit for which the maximum value of the magnetic quantum number is m = +3. The number of de Broglie waves this electron makes in one complete revolution around the nucleus is:
Show Hint
First find l from the maximum m value, then find the smallest allowed n, then use waves = n.
Work backward from the quantum number rules instead of jumping straight to Bohr's formula.
The magnetic quantum number $m$ ranges over the $2l+1$ values from $-l$ to $+l$. If the largest value of $m$ observed is $+3$, that tells us directly that $l = 3$, since $m$ can never exceed $l$.
Now use the restriction that $l$ can only take integer values from $0$ up to $n-1$. For $l = 3$ to be allowed, we need $n - 1 \geq 3$, which gives $n \geq 4$. The lowest such orbit, and hence the orbit implied here, is $n = 4$.
De Broglie combined with Bohr's model shows that a standing wave pattern exists around the orbit only if the orbit circumference equals a whole number of wavelengths, so the number of waves in the nth orbit is simply $n$.
Substituting $n = 4$ gives 4 complete waves around the nucleus in one revolution.