Which of the following is the correct plot for the probability density ψ ² (r) as a function of distance 'r' of the electron from the nucleus for 2s orbital?

Solution

Chemistry ma’am · 4 steps

Check two things: ψ² can never dip below zero, and a 2s plot must touch zero exactly once.

1. Never negative. ψ itself can be positive or negative, but ψ² is a square. So a probability density plot can never go below the r axis.
2. Count radial nodes. Radial nodes = n - l - 1. For 2s, n = 2 and l = 0, so there is 2 - 0 - 1 = 1 node. At that one distance ψ² falls exactly to zero.
3. Shape near nucleus. For every s orbital, ψ² is largest at r = 0. So the 2s curve starts high, drops to zero once, rises into a smaller hump and then fades away.

   Option	What the curve does	Times it touches zero	Fits
   (a)	Falls steadily from a maximum at the nucleus	0	1s
   (b)	Falls to zero, rises into one hump, then fades	1	2s
   (c)	Goes below the r axis	crosses once	not possible for ψ²
   (d)	Falls to zero, two humps	2	3s

4. So. Only (b) stays above the axis and has exactly one radial node, so (b) is the 2s plot.

The idea to keep. For any ns orbital, ψ² is highest at the nucleus, never negative, and drops to zero n-1 times. Count those zeros and you can read off n.

Answer from the NTA final answer key. Solution by PrepWiser.