Question

Two sets of quantum numbers with the same number of radial nodes are

• A. n = 3; l = 0; m$_l$ = 0 and n = 2; l = 0; m$_l$ = 0
• B. n = 3; l = 1; m$_l$ = 1 and n = 2; l = 1; m$_l$ = 0
• C. n = 3; l = 2; m$_l$ = 0 and n = 2; l = 1; m$_l$ = 0
• D. n = 3; l = 1; m$_l$ = -1 and n = 2; l = 1; m$_l$ = 0

Hint:

Number of radial nodes = \( n - l - 1 \). Orbitals with higher $l$ for the same $n$ have fewer radial nodes.

Answer 1

The Correct Option is C

Step 1: Formula for number of radial nodes.
Number of radial nodes = \( n - l - 1 \).
Step 2: Evaluate each option.
(A) 3s: \(3 - 0 - 1 = 2\); 2s: \(2 - 0 - 1 = 1\) → different.
(B) 3p: \(3 - 1 - 1 = 1\); 2p: \(2 - 1 - 1 = 0\) → different.
(C) 3d: \(3 - 2 - 1 = 0\); 2p: \(2 - 1 - 1 = 0\) → same (0 radial nodes).
(D) 3p: \(3 - 1 - 1 = 1\); 2p: \(2 - 1 - 1 = 0\) → different.
Step 3: Conclusion.
The pair (3d, 2p) has the same number of radial nodes (0).