For a hydrogen-like atom, the quantity asked is the probability density as a function of distance from the nucleus for the $2s$ orbital.

The radial wave function for a $2s$ orbital has the form $R_{2s}(r) \propto \left(2-\frac{r}{a_0}\right)e^{-r/2a_0}$ so $\psi^2(r) \propto \left(2-\frac{r}{a_0}\right)^2 e^{-r/a_0}$ (up to angular factors/constants).

Key features of $\psi^2(r)$ for a $2s$ orbital:

1. At $r=0$, $\psi^2(r)$ is non-zero.
2. Since the $2s$ orbital has one radial node, $\psi=0$ at one value of $r$, hence $\psi^2=0$ there.
3. After becoming zero at the node, $\psi^2$ rises again to another maximum and then decreases asymptotically toward zero as $r\to\infty$.

Therefore, the correct graph must:

• start from a non-zero value at $r=0$,
• touch the $r$-axis once (one radial node),
• rise again, and
• finally decay to zero.

Among the given options, this corresponds to Option B.

Verification with stored answer:

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