Question

8, what is the energy released in this 3" nuclear reaction u) e? (The atomic mass of 22Na is 21.994436 u and that of 22 Ne is 21.991385 Na → Ne + x 2.3315445 Mev

Answer 1

Energy during $\beta ^{+}$ decay = [mass of initial element(amu) - mass of final element(amu) - 2*(mass of electron(amu))*c²

(c² = 931 MeV)

mass of electron = 0.00054858 amu

= [21.994436 - 21.991385 - 2*0.00054858]*931 MeV

= 1.819 MeV