Answer
$${0.390 \mathrm{MeV}}$$
Work Step by Step
$$v_{\alpha f}=\frac{m_{\alpha}-m_{\mathrm{Au}}}{m_{\alpha}+m_{\mathrm{Au}}} v_{\alpha i}$$
and the recoiling gold nucleus is
$$
v_{\mathrm{Au}, f}=\frac{2 m_{\alpha}}{m_{\alpha}+m_{\mathrm{Au}}} v_{\alpha i}
$$
Therefore, the kinetic energy of the recoiling nucleus is
$${\qquad K_{\text {Auf }} =\frac{1}{2} m_{\text {Au }} v_{\text {Auf }}^{2}=\frac{1}{2} m_{\text {Au }}\left(\frac{2 m_{\alpha}}{m_{\alpha}+m_{\text {Au }}}\right)^{2} v_{\text {ai }}^{2} =K_{\text {ai }} \frac{4 m_{\text {Au }} m_{\alpha}}{\left(m_{\alpha}+m_{\text {Au }}\right)^{2}} \\ =(5.00 \mathrm{MeV}) \frac{4(197 \mathrm{u})(4.00 \mathrm{u})}{(4.00 \mathrm{u}+197 \mathrm{u})^{2}} } \\ {} {=0.390 \mathrm{MeV}}$$
