The designation of an orbital n=4 and l=0 is 4s.

Answer

To solve the question regarding the designation of an orbital with quantum numbers n=4 and l=0, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Principal Quantum Number (n)**:
- The principal quantum number (n) is given as 4. This number indicates the energy level or shell of the electron in an atom. In this case, n=4 corresponds to the fourth shell.

2. **Identify the Azimuthal Quantum Number (l)**:
- The azimuthal quantum number (l) is given as 0. This number defines the shape of the orbital. The values of l can range from 0 to (n-1). For l=0, it indicates an s orbital.

3. **Determine the Type of Orbital**:
- The value of l helps us determine the type of subshell:
- If l = 0, the subshell is s.
- If l = 1, the subshell is p.
- If l = 2, the subshell is d.
- If l = 3, the subshell is f.
- Since l=0, we conclude that it is an s subshell.

4. **Combine n and l to Designate the Orbital**:
- The designation of an orbital is represented as "n" followed by the type of subshell. Therefore, for n=4 and l=0, the designation becomes 4s.

5. **Conclusion**:
- Thus, the designation of the orbital with n=4 and l=0 is indeed 4s.

### Final Answer:
The designation of an orbital with n=4 and l=0 is **4s**.

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