Chemistry · Structure Of Atom · NEET
They are NOT the same plot, and NEET loves this trap. The ψ² curve (probability density) plots how dense the electron cloud is at a point. For 1s, ψ² is MAXIMUM right at the nucleus (r = 0) and falls off. The radial probability distribution curve plots 4πr²ψ² — the total probability in a thin spherical shell at distance r. Because it is multiplied by r², this curve is ZERO at the nucleus (r = 0 makes r² = 0) and rises to a peak a little away from the nucleus. So: ψ² is max at nucleus, but 4πr²ψ² is zero at nucleus. Always check which curve the question shows.
The radial probability is 4πr²ψ². At the exact nucleus r = 0, so the r² factor makes the whole thing 0, even though ψ² itself is large there. Physically: a shell of zero radius has zero volume, so it can hold no electron. As r grows, the shell volume (4πr²) grows fast while ψ² shrinks slowly at first, so the probability rises to a peak, then falls as ψ² finally drops off.
A radial node is a value of r where the curve touches zero BETWEEN the nucleus and infinity (not counting r = 0 or r = ∞). The number of radial nodes = n − l − 1. For 1s: n = 1, l = 0, so 1 − 0 − 1 = 0 radial nodes — the curve has one smooth hump, no dip to zero. For 2s: n = 2, l = 0, so 2 − 0 − 1 = 1 radial node — the curve rises, drops all the way to zero (the node), then rises to a second, bigger hump. That is why 2s has TWO peaks and 1s has ONE.
Count how many times the curve CROSSES the horizontal axis (touches zero) between the nucleus and infinity, ignoring r = 0 and r = ∞. Each crossing is one radial node. 1s = 0 crossings, 2s = 1 crossing, 3s = 2 crossings. This is exactly what ReNEET 2026 tested: the plot that crossed zero twice was 3s (2 radial nodes). Formula check: radial nodes = n − l − 1.
The highest peak of the 1s radial probability curve sits at r = a₀ = 0.529 Å, the Bohr radius. This is the single most probable distance to find the 1s electron. For 2s, the tallest (outer) peak lies farther out, which is why 2s is a bigger orbital than 1s. Remember the size order: 1s < 2s < 3s.
No. The 2s curve has a small inner peak close to the nucleus and a larger outer peak. The electron has some probability of being found in BOTH regions, separated by the node where probability is exactly zero. The outer peak is bigger, so the electron is more likely to be found there, but the inner peak (penetration near the nucleus) is why 2s is lower in energy than 2p in multi-electron atoms.
Consider the schematic plots of the orbital wavefunction ψ_r against distance r from the nucleus (A: monotonically decreasing; B: one sign-change; C: two ripples crossing zero twice; D: oscillatory). The figure representing two radial nodes in the orbital is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. The 1s curve has one peak and no node. The 2s curve has two peaks with one radial node (a point where probability drops to zero) between them.
One. Using radial nodes = n − l − 1, for 2s we get 2 − 0 − 1 = 1. This shows up as the single dip to zero on the curve.
4πr² is the surface area of a sphere of radius r. Multiplying it by ψ² gives the total probability in a thin spherical shell at that distance, which is what we actually measure moving outward from the nucleus.
At r = 0.529 Å, the Bohr radius (a₀). This is the position of the single peak of the 1s radial probability curve for hydrogen.
The small inner peak of 2s lets the electron penetrate close to the nucleus, so it feels more nuclear pull. This extra penetration lowers 2s energy compared to 2p in multi-electron atoms.
