Consider Xenon `(Z=54)`. The maximum number of electrons in this atom that have the values for their quantum numbers as `n=4, l=3` and `s=1/2` in its ground state is:

Options

A. Zero
B. `7`
C. `9`
D. `14`

Answer

To solve the problem, we need to determine the maximum number of electrons in the ground state of Xenon (Xe, Z=54) that can have the quantum numbers \( n=4 \), \( l=3 \), and \( s=\frac{1}{2} \).

### Step-by-Step Solution:

1. **Understanding Quantum Numbers**:
- The principal quantum number \( n \) indicates the energy level of the electron. Here, \( n=4 \) means we are looking at the fourth energy level.
- The azimuthal quantum number \( l \) indicates the subshell. For \( l=3 \), this corresponds to the f subshell (since \( s=0, p=1, d=2, f=3 \)).
- The spin quantum number \( s=\frac{1}{2} \) indicates the spin of the electron.

2. **Identify the Relevant Subshell**:
- For \( n=4 \) and \( l=3 \), we are looking at the 4f subshell.

3. **Determine the Electron Capacity of the 4f Subshell**:
- The f subshell can hold a maximum of 14 electrons (since each f subshell can hold \( 2(2l + 1) = 2(2*3 + 1) = 14 \) electrons).

4. **Check the Electron Configuration of Xenon**:
- The electron configuration of Xenon is:
\[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^6 \, 5s^2 \, 4d^{10} \, 5p^6 \]
- In this configuration, the 4f subshell is not filled because Xenon only has 54 electrons, and the 4f subshell starts filling after Xenon, in the Lanthanides series.

5. **Conclusion**:
- Since the 4f subshell is not occupied in the ground state of Xenon, the maximum number of electrons that can have the quantum numbers \( n=4 \), \( l=3 \), and \( s=\frac{1}{2} \) is **0**.

### Final Answer:
The maximum number of electrons in Xenon that have the quantum numbers \( n=4 \), \( l=3 \), and \( s=\frac{1}{2} \) is **0**.