A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is λ. The electric field at the center of this arc is:
a. λ / (2π ε₀ a)
b. λ / (2π ε₀ a²)
c. λ / (4π² ε₀ a)
d. λ² / (2π ε₀ a)

To find the electric field at the center of a semi-circular arc of radius \(a\) uniformly charged with linear charge density \(\lambda\), we can proceed as follows:

### Setup
- The charge per unit length is \(\lambda\).
- The total charge on the semi-circular arc is \(Q = \lambda \times \) arc length \(= \lambda \times \pi a\).
- We are interested in finding the electric field \( \mathbf{E} \) at the center of the semi-circle.

### Electric Field Contribution
- Consider a small element \(d\theta\) on the arc at an angle \(\theta\) from the horizontal axis.
- The length of the element is \(dl = a \, d\theta\).
- The charge on this element is \(dq = \lambda \, dl = \lambda a \, d\theta\).

The electric field due to this small element at the center is:

\[
dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{a^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda a \, d\theta}{a^2} = \frac{\lambda}{4 \pi \varepsilon_0 a} d\theta
\]

### Symmetry Consideration
- The vertical components of the electric field due to symmetrically opposite elements on the arc will cancel out.
- The horizontal components will add up.

The horizontal component of the electric field is:

\[
dE_x = dE \cos \theta = \frac{\lambda}{4 \pi \varepsilon_0 a} \cos \theta \, d\theta
\]

### Total Electric Field
To find the total electric field, integrate \(dE_x\) over the entire semi-circular arc from \(-\pi/2\) to \(\pi/2\):

\[
E = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4 \pi \varepsilon_0 a} \cos \theta \, d\theta
\]

The integral of \(\cos \theta\) over \(-\pi/2\) to \(\pi/2\) is:

\[
\int_{-\pi/2}^{\pi/2} \cos \theta \, d\theta = 2
\]

Thus, the electric field at the center is:

\[
E = \frac{\lambda}{4 \pi \varepsilon_0 a} \times 2 = \frac{\lambda}{2 \pi \varepsilon_0 a}
\]

### Final Answer
The correct answer is:

\[
\boxed{ \frac{\lambda}{2 \pi \varepsilon_0 a} }
\]

This matches with option **(a)**.