Radial Probability Distribution
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?
Select the correct option:
Solution
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.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- radial probability distribution
- Difficulty
- Hard
- Year
- 2025
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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