Questions: Hydrogen Atom Solution: Radial Wavefunction

Question 1

The 3s orbital has n=3, l=0 and the 3p orbital has n=3, l=1. How many radial nodes does each have, and what does the difference tell us?

• A. 3s has 3 nodes, 3p has 3 nodes — principal quantum number alone determines node count
• B. 3s has 2 nodes, 3p has 1 node — greater angular momentum replaces radial nodes with angular nodes
• C. 3s has 1 node, 3p has 2 nodes — angular orbitals have more complex radial structure
• D. 3s has 0 nodes, 3p has 1 node — s orbitals are nodeless because they are spherically symmetric

Answer: (B)

Radial nodes = n − l − 1. For 3s: 3 − 0 − 1 = 2 radial nodes. For 3p: 3 − 1 − 1 = 1 radial node. The pattern reveals a conservation of total nodes: both have n − 1 = 2 total nodes, but 3p trades a radial node for an angular node (the nodal plane through the nucleus). This trade-off has a physical consequence: the extra inner lobe in 3s gives it greater nuclear penetration, which matters enormously in multi-electron atoms where s electrons experience higher effective nuclear charge.

Question 2

The radial wavefunction R(r) of the 1s orbital is largest at r = 0 (right at the nucleus). Yet the radial probability density P(r) peaks at r = a₀, not r = 0. Why?

• A. The 1s wavefunction has a node at r = 0 that cancels the probability density
• B. The Pauli exclusion principle prevents the electron from being at the nucleus
• C. P(r) = r²|R(r)|² includes an r² factor from spherical volume elements — at r = 0 this factor is zero, so even a large R(r) gives zero probability
• D. The electron's kinetic energy diverges at r = 0, making that region classically forbidden

Answer: (C)

The probability of finding the electron in a thin shell of thickness dr at radius r is P(r) dr = r²|R(r)|² dr. The r² comes from the volume element 4πr² dr in spherical coordinates — there is simply more volume in a shell at large r than at small r. Even though |R(r)|² is largest at the origin, the r² factor is zero there, so P(0) = 0. Moving outward, the r² factor grows while R(r) decreases exponentially; the product peaks at r = a₀ for the 1s orbital. This is why the Bohr model's prediction of r = a₀ emerges from the correct quantum mechanical treatment.

Question 3

The radial probability density P(r) for the 1s orbital peaks at a distance from the nucleus, not at r = 0, even though the radial wavefunction R(r) is largest at r = 0.

• T. True
• F. False

Answer: True

True. P(r) = r²|R(r)|² incorporates a volume-element factor r² that goes to zero at the origin. This means that despite R(1s) being maximum at r = 0, the actual probability of finding the electron in a thin spherical shell peaks at r = a₀. The distinction between |R(r)|² (probability density per unit volume) and P(r) (probability per unit radial distance) is essential for correctly interpreting orbital structure.

Question 4

Going from an s orbital to a p orbital of the same principal quantum number (e.g., 2s → 2p) generally increases the number of radial nodes.

• T. True
• F. False

Answer: False

False — it decreases radial nodes. Radial nodes = n − l − 1, so higher angular momentum l means fewer radial nodes. For n = 2: the 2s has 2 − 0 − 1 = 1 radial node, while the 2p has 2 − 1 − 1 = 0 radial nodes. The total node count (radial + angular) stays fixed at n − 1 = 1 for both; the 2p trades its radial node for an angular nodal plane. A common misconception is that more complex orbitals must have more nodes overall — in fact, total nodes depend only on n.

Question 5

Why do 2s and 2p electrons in a multi-electron atom experience different effective nuclear charges, even though both have n = 2 and both have zero radial nodes — wait, that's not right. Explain the actual difference in nuclear penetration between 2s and 2p orbitals.

Model answer: The 2s orbital has one radial node and a small but nonzero inner lobe of electron density close to the nucleus, giving 2s electrons significant probability of being found near the nucleus. The 2p orbital has no radial nodes but has an angular nodal plane through the nucleus, with zero probability at r = 0 and a radial probability distribution that stays farther from the nucleus. The inner lobe of 2s allows it to 'penetrate' past the shielding electrons to feel a higher effective nuclear charge, making 2s lower in energy than 2p in multi-electron atoms (breaking the hydrogen-like n = 2 degeneracy).

In hydrogen, 2s and 2p are degenerate — they have the same energy. In multi-electron atoms, inner electrons shield the outer electrons from the full nuclear charge. But penetration (probability of being close to the nucleus) varies with orbital shape. The inner lobe of the 2s radial probability distribution gives 2s electrons a higher effective nuclear charge than 2p electrons, lowering the 2s energy relative to 2p. This penetration-and-shielding argument explains the Aufbau filling order and why, for example, 4s fills before 3d.