Answer
a) $^{238}_{92}U\rightarrow ^{234}_{90}Th+^{4}_{2}He$
b) $m_{^{234}_{90}Th}=234.04u$
Work Step by Step
a) An $\alpha$ particle consists of 2 protons and 2 neutrons, so $^{238}_{92}U\rightarrow ^{234}_{90}Th+^{4}_{2}He$
b) $KE=[m_{^{238}_{92}U}-m_{^{234}_{90}Th}-m_{^{4}_{2}He}]c^2$
$m_{^{234}_{90}Th}=234.04u$
