Answer
(a) $^{35}_{16}S\rightarrow ^{35}_{17}Cl+e^{-};168KeV$
(b) $^{212}_{82}Pb\rightarrow ^{212}_{83}Bi+e^{-};566KeV$
Work Step by Step
(a) The required decay process is given as
$^{35}_{16}S\rightarrow ^{35}_{17}Cl+e^{-}$
$\Delta m=m_f-m_i$
$\Delta m=34.968853u-34.969033u=-0.00018u$
Now the energy released is given as
$E=|\Delta m|c^2$
We plug in the known values to obtain:
$E=(0.00018u)(\frac{931.494\frac{MeV}{c^2}}{1u})c^2$
$\implies E=168KeV$
(b) We know that the required decay process is given as
$^{212}_{82}Pb\rightarrow ^{212}_{83}Bi+e^{-}$
$\Delta m=m_f-m_i$
$\Delta m=211.991272u-211.99188u=-0.000608u$
Now the energy released can be determined as
$E=|\Delta m|c^2$
$\implies E=|-(0.000608u)(\frac{931.494\frac{MeV}{c^2}}{1u})|c^2$
$\implies E=566KeV$
