Answer
(a) $^{18}_{9}F\rightarrow ^{18}_{8}O+e^{+}+\nu;0.634MeV$
(b) $^{22}_{11}Na\rightarrow ^{22}_{10}Ne+e^{+}+\nu;1.819MeV$
Work Step by Step
(a) We know that the required decay process is given as
$^{18}_{9}F\rightarrow ^{18}_{8}O+e^{+}+\nu$
$\Delta m=m_i-m_f$
$\Delta m=18.000938-17.9997076=0.0012304u$
$\implies \Delta m=0.0012304u(\frac{931.5MeV/c^2}{1u})=1.1461176MeV/c^2$
Now, the energy released can be determined as
$E=|\Delta m|c^2$
We plug in the known values to obtain:
$E=|1.1461176 MeV/c^2|c^2=0.634MeV$
(b) The required decay process is given as
$^{22}_{11}Na\rightarrow ^{22}_{10}Ne+e^{+}+\nu$
$\Delta m=21.994435u-21.9924182u=0.0019538u$
$\implies \Delta m=0.0019538u(\frac{931.5MeV/c^2}{1u})=1.8199647MeV/c^2$
Now the energy released can be determined as
$E=|1.8199647MeV/c^2|c^2=1.8199647MeV\approx 1.819MeV$
