An orbital has only positive values of wave function at all distances from the nucleus. Find the value of `(n+l)` for this orbital.

Explanation

To solve the problem, we need to determine the value of \( n + l \) for an orbital that has only positive values of the wave function at all distances from the nucleus.

### Step-by-Step Solution:

1. **Understanding the Wave Function**: The wave function \( \psi \) describes the probability amplitude of finding an electron in a given region of space. If the wave function is always positive, it indicates that there are no nodes (points where the wave function is zero) in the orbital.

2. **Identifying the Orbital**: The only orbital that has a wave function that is always positive and does not touch or cross the zero line is the 1s orbital. This is because the 1s orbital has a spherical shape and does not have any nodes.

3. **Quantum Numbers**:
- The principal quantum number \( n \) for the 1s orbital is 1.
- The azimuthal quantum number \( l \) for the 1s orbital is 0 (since s orbitals correspond to \( l = 0 \)).

4. **Calculating \( n + l \)**:
\[ n + l = 1 + 0 = 1 \]

5. **Conclusion**: Therefore, the value of \( n + l \) for the orbital with only positive values of the wave function is 1.

### Final Answer:
The value of \( n + l \) for this orbital is **1**.