Answer
237.0483u.
Work Step by Step
Energy is conserved. The rest energy of the starting americium nucleus is equal to the rest energies of the 2 neptunium nucleus and the alpha particle, plus the kinetic energy of the alpha particle. We neglect the KE of the neptunium nucleus.
$$m_{Am}c^2=m_{Np}c^2+m_{\alpha}c^2+KE_{\alpha}$$
$$ m_{Np}= m_{Am}-m_{\alpha}-\frac{KE_{\alpha}}{c^2}$$
$$=241.05682u-4.00260u-\frac{5.5MeV}{c^2}(\frac{1u}{931.49MeV/c^2})=237.0483u$$
