Question

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:

• A. 4
• B. 5
• C. 2
• D. 0

Hint:

First find l from the maximum m value, then find the smallest allowed n, then use waves = n.

Answer 1

The Correct Option is A

Step 1: Recall the rule connecting the magnetic quantum number m to the azimuthal quantum number l. For a given l, m can take values from -l to +l, including zero. So if the maximum value of m is +3, then l must be 3, because the highest value m can reach is l itself.

Step 2: Recall the rule connecting l to the principal quantum number n. The azimuthal quantum number l can range from 0 to (n - 1), so l can never equal or exceed n. Since l = 3, the smallest n that allows l = 3 is n = 4, because for n = 4, l can be 0, 1, 2 or 3.

Step 3: Use the Bohr postulate connecting orbit number to de Broglie waves. Bohr's quantisation condition states that the circumference of a stationary orbit must be a whole number of de Broglie wavelengths, written as \( 2\pi r_n = n\lambda \). This means the electron completes exactly n full standing waves as it goes once around the nucleus in the nth orbit.

Step 4: Since the orbit here corresponds to n = 4, the electron makes 4 complete de Broglie waves in one revolution. This matches option A, which is the given answer.

Note: this question uses m = +3 as a marker for l = 3, and expects the student to read off the orbit number n = 4 directly, which is exactly what is done above.