Answer
(a) $^{212}_{84}P_{\circ}\rightarrow ^{208}_{82}Pb+^4_2He$; $8.95MeV$
(b) $^{239}_{94}Pu\rightarrow ^{235}_{92}U+^4_2He;$ $5.244MeV$
Work Step by Step
(a) It is given that $^{212}_{84}P_{\circ}$ decays and emits a particle ($^4_2He$)
Thus, the required process is $^{212}_{84}P_{\circ}\rightarrow ^{208}_{82}Pb+^4_2He$
Now $\Delta m=m_f-m_i$
$\Delta m=211.97924u-211.988852u=-0.009612u$
The required energy released is given as
$E=|\Delta m|c^2$
We plug in the known values to obtain:
$E=(0.009614u)|\frac{93.1494\frac{MeV}{c^2}}{1u}|c^2$
$\implies E=8.95MeV$
(b) The required reaction is
$^{239}_{94}Pu\rightarrow ^{235}_{92}U+^4_2He$
Now $\Delta m=m_f-m_i$
$\Delta m=239.046528u-239.052158u=-0.00563$
The energy released is given as
$E=|\Delta m|c^2$
We plug in the known values to obtain:
$E=|-(0.00563u)(\frac{931.494\frac{MeV}{c^2}}{1u})|c^2=5.244MeV$
