Answer
(a) $E = 4.27 MeV$
(b) $v= 2.43 \times 10^5 m/s$
Work Step by Step
(a) The energy released is the same as the mass defect, which is the total difference between initial mass and final mass of substances.
$E = (238.050788 u – 234.043601 u – 4.002603 u )931.5 MeV/u$
$E = (0.004584 u)931.5 MeV/u$
$E = 4.27 MeV$
(b) Taking the ratio of both kinetic energies, $K = \frac{p^2}{2m} $
$\frac{K_{Th}}{K_{\alpha}} = \frac{\frac{p^2}{2m_{Th}}}{\frac{p^2}{2m_{\alpha}}}$
$\frac{K_{Th}}{K_{\alpha}} = \frac{p^2}{2m_{Th}} \times \frac{2m_{\alpha} }{p^2}$
$\frac{K_{Th}}{K_{\alpha}} = \frac{m_{\alpha}}{m_{Th}}$
$\frac{K_{Th}}{K_{\alpha}} = \frac{4}{234} $
$K_{Th} =\frac{4}{234+4} (K_{\alpha})$
Know that $K_{\alpha} = k_{total}$
$K_{Th} =\frac{4}{238} (4.27 MeV)$
$K_{Th} = 0.07176 MeV = 1.148 \times 10^{-14} J$
From the kinetic energy, solve for velovity.
$E_K = \frac{1}{2} mv^2$
$v= \sqrt{\frac{2E_K}{m}}$
$v= \sqrt{\frac{(2)(1.148 \times 10^{-14} J) }{(234.043601)(1.6605 \times 10^{-27} kg)}}$
$v= 2.43 \times 10^5 m/s$
