Answer
$$4.61 \mathrm{MeV} $$
Work Step by Step
final kinetic energy of alpha particle is
$\begin{aligned} K_{\alpha f} &=\frac{1}{2} m_{\alpha} v_{\alpha f}^{2}=\frac{1}{2} m_{\alpha}\left(\frac{m_{\alpha}-m_{\mathrm{Au}}}{m_{\alpha}+m_{\mathrm{Au}}}\right)^{2} v_{\alpha i}^{2}=K_{\alpha i}\left(\frac{m_{\alpha}-m_{\mathrm{Au}}}{m_{\alpha}+m_{\mathrm{Au}}}\right)^{2} \\ &=(5.00 \mathrm{MeV})\left(\frac{4.00 \mathrm{u}-197 \mathrm{u}}{4.00 \mathrm{u}+197 \mathrm{u}}\right)^{2} \\ &=4.61 \mathrm{MeV} \end{aligned}$
