Question:
Which of the following statement is incorrect?}
Which of the following statement is incorrect?}
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For \(1s\) orbital, probability density is highest at the nucleus and then decreases outward.
Updated On: Apr 24, 2026
- The square of the wave function at a point gives the probability density of the electron at that point.
- For 1s orbital the probability density is maximum at the nucleus and it increases sharply as we move away from it.
- For 2s orbital the probability density first decreases sharply to zero and again starts increasing.
- Node is the region where the probability density function of electron reduces to zero.
- The number of nodes for 3s orbital is 2.
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The Correct Option is B
Solution and Explanation
For the \(1s\) orbital, the probability density is maximum at the nucleus, but it does not increase as we move away from the nucleus.
Instead, it decreases continuously with distance.
Now check the statements:
Hence, the incorrect statement is: \[ \boxed{(B)} \]
- (A) is correct because \(\psi^2\) gives probability density.
- (B) is incorrect because the probability density for \(1s\) decreases, not increases, away from the nucleus.
- (C) is correct for \(2s\), which has a node.
- (D) is correct because node means zero probability density.
- (E) is correct because for \(3s\), total nodes \(= n-1 = 2\).
Hence, the incorrect statement is: \[ \boxed{(B)} \]
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