Questions: Radial Wavefunctions and Probability Distributions in Hydrogen

Question 1

For the hydrogen 1s orbital, the wavefunction amplitude |R(r)|² is actually largest at r = 0 — right at the nucleus. Yet the most probable location for the electron is at r = a₀ ≈ 0.53 Å. Why aren't these contradictory?

• A. The 1s wavefunction has a node at r = 0, making |R(r)|² zero there
• B. The probability of finding the electron in a shell at radius r is proportional to r²|R(r)|², which is zero at r = 0 because the spherical shell has zero area there
• C. The nucleus electrostatically repels the electron, pushing probability density outward
• D. There is a normalization convention that sets the wavefunction to zero at the origin

Answer: (B)

The probability of finding the electron in a thin shell between r and r + dr is the radial probability density P(r) = r²|R(r)|² times dr — not |R(r)|² times dr. The r² factor comes from the surface area of the spherical shell (4πr²). At r = 0, the shell has zero area, so P(r) = 0 regardless of how large |R(r)|² is there. The peak of P(r) is pushed outward to a₀, where the product of large wavefunction amplitude and large shell area is maximized. This is purely a geometric effect, not physics that 'pushes' the electron away from the nucleus.

Question 2

The 2s and 2p orbitals have the same principal quantum number (n = 2), but 2s electrons can 'penetrate' closer to the nucleus than 2p electrons. What accounts for this difference?

• A. 2s electrons have higher energy than 2p electrons, giving them more kinetic energy to overcome nuclear repulsion
• B. The angular momentum quantum number ℓ creates a centrifugal-like barrier term ℓ(ℓ+1)/r² in the effective radial potential that suppresses near-nucleus probability for 2p electrons
• C. The 2s orbital has more radial nodes than 2p, forcing its probability outward toward larger r
• D. 2p electrons are heavier due to carrying angular momentum, increasing their effective mass near the nucleus

Answer: (B)

In the effective radial potential, angular momentum contributes a term ℓ(ℓ+1)ℏ²/(2mr²) that acts like a centrifugal barrier, repelling radial probability away from r = 0 for any ℓ > 0. For 2p (ℓ=1), this barrier is nonzero, suppressing probability near the nucleus. For 2s (ℓ=0), there is no such barrier, so the 2s wavefunction can reach the nucleus. This penetration means 2s electrons experience more of the nuclear charge (less shielding by inner electrons), which lowers their energy below 2p in multi-electron atoms — driving the energy-level splitting that underlies the periodic table.

Question 3

The Bohr radius a₀ ≈ 0.53 Å is the most probable electron-nucleus distance for most hydrogen orbitals (most values of n and ℓ).

• T. True
• F. False

Answer: False

The Bohr radius a₀ is the most probable radius only for the 1s (n=1, ℓ=0) orbital, where the radial probability density P(r) = r²|R(r)|² peaks at exactly a₀. For excited states (higher n), the peak of P(r) moves to larger radii — roughly n²a₀ for states with ℓ = n−1. This is why electrons in higher shells are farther from the nucleus, have lower ionization energies, and are more available for chemical bonding. The Bohr model accidentally got the most probable radius right for the ground state.

Question 4

For the 1s hydrogen orbital, the radial probability density P(r) = r²|R(r)|² equals zero at r = 0, even though the wavefunction amplitude |R(r)|² is nonzero there.

• T. True
• F. False

Answer: True

At r = 0, the factor r² = 0, so P(r) = r² × |R(r)|² = 0, regardless of the nonzero value of |R(r)|². The physical interpretation: the probability of finding the electron in a shell of vanishing thickness at r = 0 is zero because the shell has zero volume. This is purely geometric. The wavefunction *amplitude* is nonzero at the nucleus (s orbitals have nonzero electron density at the nucleus, which matters for hyperfine structure in atomic physics), but the *probability of finding the electron in a shell* there is zero.

Question 5

Explain why the most probable radius for the hydrogen 1s electron is NOT where the wavefunction amplitude |R(r)|² is largest, and what quantity must instead be maximized to find the most probable radius.

Model answer: The probability of finding the electron in a thin shell at radius r is proportional to the shell's volume, which is 4πr²dr. The relevant quantity is the radial probability density P(r) = r²|R(r)|², not |R(r)|² alone. For 1s, |R(r)|² is maximum at r = 0 (the wavefunction peaks at the nucleus), but P(r) = 0 there because the shell area r² = 0. The most probable radius is where P(r) — the product of large wavefunction amplitude and large shell area — is maximized. For 1s, this occurs at r = a₀, the Bohr radius.

This is one of the most important distinctions in quantum mechanics and is often missed by students familiar with the Bohr model. The Bohr model treats the electron as orbiting at a fixed radius; quantum mechanics gives a probability distribution. The most probable radius is a property of the radial *probability density* P(r), not the wavefunction itself. The r² geometric factor is what causes these to differ and is why even for the simplest orbital, finding the electron requires integrating over spherical shells rather than reading off the wavefunction amplitude.