Question

What is the energy released in this β + nuclear reaction 29 16 S → 29 15 P + 0 1 e ? (The atomic mass of 29 S is 28.996615 u and that of 29 P is 28.9818 u)

13.3 MeV is incorrect

Answer 1

²⁹S₁₆²⁹P₁₅ + ⁰e₁

Atomic numbers before reaction = 16

Atomic numbers after reaction = 15 + 1 = 16

Mass numbers before reaction = 29

Mass numbers after reaction = 29 + 0 = 29

{ Atomic numbers and mass numbers are balanced }

Total mass before the reaction, m_initial = 28.996615 u

Total mass after the reaction, m_final = (28.9818 u) + (0.00054858 u) = 28.98234858 u

Mass defect is given by -

m = m_initial - m_final

m = [(28.996615 u) - (28.98234858 u)]

m = 0.01426642 u

Converting u into kg :

m = (0.01426642 u) [(1.66054 x 10^-27 kg) / (1 u)]

m = 2.3689 x 10^-29 kg

Therefore, the energy released in this ⁺ nuclear reaction which will be given by -

E = m c²

E = [(2.3689 x 10^-29 kg) (3 x 10⁸ m/s)²]

E = 2.13201 x 10^-12 J

Converting J into MeV :

E = (2.13201 x 10^-12 J) [(6241506479963.2 MeV) / (1 J)]

E = 13.3 MeV