Question

The atomic mass of ₇N¹⁵ is 15.000108 a.m.u. and that of ₈O¹⁶ is 15.994915 a.m.u. If the mass of a proton is 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is
1. 0.013018 MeV
2. 12.13 MeV
3. 13.018 MeV
4. 12.13 eV

Answer 1

Correct Answer - Option 2 : 12.13 MeV

Concept:-

The energy is required to remove the least tightly bound proton will be equal to the mass difference in products and reactants and products in the same units and then multiplying by c² (931.5) MeV. It is also called the Q-Value for the given nuclear reaction.

Energy + ₈O¹⁶ \(\longrightarrow\) ₇N¹⁵ + ₁H¹

Calculation:-

Given: Mass of a proton (MH) = 1.007825 a.m.u.

The atomic mass of 7N15 (MN) = 15.000108 a.m.u. and 

The atomic mass of 8O16 (M0) = 15.994915 a.m.u.

Energy or (Q) = [(M_N + M_H) – M₀] c²

             = [(15.000108 + 1.007825) – 15.994915] × 931.5 MeV

             = 12.13 MeV

Additional points:-

•  A nuclear reaction for which the Q value is positive is called exothermic while endothermic for the negative value.
• The value of 1c² = 931.5 MeV/a.m.u.