Step 1: Understanding the Concept:
The radial probability density function, \(4\pi r^2 R(r)^2\), describes the probability of finding an electron in a thin spherical shell of radius \(r\) and thickness \(dr\).
A crucial feature of these plots is the "radial nodes"—values of \(r\) (excluding zero and infinity) where the probability density drops to zero.
The number of these nodes is characteristic of the orbital's principal and azimuthal quantum numbers.
Step 2: Key Formula or Approach:
The total number of radial nodes for any orbital is calculated using the formula:
\[ \text{Radial Nodes} = n - l - 1 \]
Where:
- \(n\) = Principal quantum number.
- \(l\) = Azimuthal quantum number (for a \(p\) orbital, \(l = 1\)).
Step 3: Detailed Explanation:
By examining the provided plot (C54.png):
1. **Counting Nodes:** The graph starts at the origin (\(r=0\)), goes through a peak, descends to touch the horizontal axis (zero probability), and then rises to a second, larger peak before tailing off.
2. The point where the curve touches the zero-line between the two peaks represents one radial node.
3. **Applying the Formula:** We are given that the electron is in an \(np\) orbital.
- For a \(p\) orbital, the azimuthal quantum number \(l = 1\).
- From the graph, Radial Nodes = 1.
- Substitute these into the formula:
\[ 1 = n - 1 - 1 \]
\[ 1 = n - 2 \]
\[ n = 3 \]
Thus, the orbital is \(3p\).
Additionally, the number of peaks in such a plot is given by \(n - l\).
Since the graph shows 2 peaks:
\[ 2 = n - 1 \Rightarrow n = 3 \]
Both the node count and peak count confirm that \(n = 3\).
Step 4: Final Answer:
The principal quantum number \(n\) is 3. The correct option is (B).