Alpha particles of energy 5.0 MeV pass a gold nucleus, with Z = 79…

Step 1. Closest approach. $r_0 = \dfrac{2 \times 79 \times 1.44\ \text{MeV fm}}{5.0\ \text{MeV}} = 45.5\ \text{fm}$.

Step 2. The angle. $\tan\dfrac{\theta}{2} = \dfrac{45.5\ \text{fm}}{2 \times 12\ \text{fm}} = 1.90$. So $\dfrac{\theta}{2} = 62.2^\circ$ and $\theta = 124^\circ$.

Step 3. A deflection of $90^\circ$. Then $\tan 45^\circ = 1$, so $b = \dfrac{r_0}{2} = 22.8\ \text{fm}$.