We are asked to identify an orbital from the graph of $|\psi|^2$ versus radial distance $r$.

Key idea:

• The number of radial nodes helps identify the orbital.
• For a hydrogen-like orbital: $\text{number of radial nodes} = n - l - 1$ where:
• $n$ = principal quantum number
• $l$ = azimuthal quantum number

Now check each option:

A: $3s$ orbital

• For $3s$, $n=3$, $l=0$
• Radial nodes: $3-0-1 = 2$
• So its $|\psi|^2$ vs $r$ graph should show two radial nodes.

B: $2s$ orbital

• For $2s$, $n=2$, $l=0$
• Radial nodes: $2-0-1 = 1$
• So its graph should show one radial node.

C: $2p$ orbital

• For $2p$, $n=2$, $l=1$
• Radial nodes: $2-1-1 = 0$
• So it has no radial node.
• Also, for a $p$ orbital, $\psi=0$ at $r=0$.

D: $1s$ orbital

• For $1s$, $n=1$, $l=0$
• Radial nodes: $1-0-1 = 0$
• So it has no radial node.
• Also, $|\psi|^2$ is maximum at $r=0$ and then decreases smoothly.

From the given graph, the wave function probability density $|\psi|^2$ shows one radial node (it becomes zero once and then rises again).

Therefore, the orbital represented is: $\boxed{2s\text{ orbital}}$

Option-wise conclusion:

• A: Incorrect
• B: Correct
• C: Incorrect
• D: Incorrect

Comparison with stored correct answer:

• Stored correct answer = B
• Derived answer = B
• Hence, they agree.