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Radial Probability Distribution

Hardchemistry

The radial probability distribution function 4πr^2 ψ^2 for a 2s orbital in hydrogen shows two maxima. What is the physical significance of the local minimum (radial node) that appears between these two maxima?

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About This Question

Subject
chemistry
Chapter
atomic structure
Topic
radial probability distribution
Difficulty
Hard
Year
2025
Tags
radial probability distributionradial nodes2s orbitalquantum tunnelingwavefunction interpretationprobability density

Solution

Correct Answer:

The probability of finding the electron at exactly this radial distance is zero, but the electron can exist on either side of this spherical surface

The radial probability distribution P(r) = 4πr^2 ψ^2 gives the probability of finding an electron in a thin spherical shell of radius r and thickness dr. A radial node occurs at a specific value of r where the radial wavefunction R(r) passes through zero, making both ψ = 0 and P(r) = 0 at that point. For the 2s orbital (n=2, l=0), there is one radial node (n-l-1 = 2-0-1 = 1) at r ≈ 2a_0. Crucially, the zero probability at the node means the electron has zero chance of being found at that exact radial distance, yet the electron can exist both closer (inner maximum near r ≈ a_0/2) and farther (outer maximum near r ≈ 5a_0) from the nucleus. Quantum mechanically, the electron transitions between these regions through quantum tunneling across the zero-probability surface without actually passing through it in the classical sense—this is the quantum tunneling behavior that has no classical analogue. Option (A) is incorrect; the electron does not 'rest' anywhere; velocity and position are governed by the uncertainty principle. Option (C) is incorrect; spin state is independent of radial position. Option (D) is incorrect; a node does not split an orbital into two separate orbitals—it remains one orbital with two probability lobes. This concept distinguishes quantum mechanical behavior from classical orbital models and is a high-level JEE Advanced topic. Plausibility check: the 2s orbital has exactly n-l-1=1 radial node, consistent with the single minimum described; the total 4πr^2ψ^2 integrated over all space still equals 1, confirming the electron exists as a whole in one orbital.

This hard difficulty chemistry question is from the chapter atomic structure, covering the topic of radial probability distribution. It appeared in the 2025 exam.

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