The given problem states that a and b are two unit vectors, and we need to determine the angle between them when √3a - b is also a unit vector. Let's break down the problem step by step:

1. Understanding Unit Vectors:
- A unit vector is a vector with a magnitude of 1. It represents the direction of a vector without considering its length.
- In other words, a unit vector is a vector divided by its magnitude.

2. Analyzing √3a - b:
- We are given that √3a - b is a unit vector. Let's denote it as c.
- So, c = √3a - b.

3. Finding the Magnitude of c:
- The magnitude of a vector is the length or size of the vector.
- Since c is a unit vector, its magnitude is 1.
- Therefore, |c| = 1.

4. Expanding c:
- Using the distributive property, we can expand c as follows:
c = √3a - b
c = √3a + (-1)b
c = √3a + (-1)b

5. Computing the Magnitude of c:
- To find the magnitude of c, we square each component, sum them, and take the square root.
- |c|^2 = (√3)^2|a|^2 + (-1)^2|b|^2 + 2(√3)(-1)|a||b|cosθ
|c|^2 = 3|a|^2 + |b|^2 - 2√3|a||b|cosθ

6. Substituting the Values:
- Since a and b are unit vectors, |a| = |b| = 1.
- |c|^2 = 3 + 1 - 2√3cosθ
- |c|^2 = 4 - 2√3cosθ

7. Simplifying |c|^2:
- As |c| = 1, we can write the equation as:
1 = 4 - 2√3cosθ
- Rearranging the terms, we have:
2√3cosθ = 3
cosθ = 3/(2√3)
cosθ = √3/2

8. Finding the Angle θ:
- From the equation cosθ = √3/2, we can determine the angle θ by taking the inverse cosine (arccos) of both sides.
- θ = arccos(√3/2)
- Using a calculator, we find θ ≈ 30°.

9. Final Answer:
- The angle between vectors a and b is approximately 30° when √3a - b is a unit vector.

By following the above steps, we have determined the angle between the two unit vectors a and b when the given condition is satisfied.