Consider the nuclear reaction $ ^{14} _{7} N + ^{4} _{2} He → ^{17} _{8} O + ^{1} _{1} H$. The known atomic masses are $^{14} _{7} N: 14.003074 u\ ^{4} _{2} He: 4.002603 u\ ^{1} _{1} H: 1.007825 u\ ^{17} _{8} O$: 16.999131 u What is the Q-value (or reaction energy) of this reaction? (1 u = 931.5 MeV/c²)

Question

Quizplus · Accepted Answer

The Q-value is given by Q = (mass of reactants) - (mass of products) 
= [(14.003074 u) + (4.002603 u)] - [(17.999131 u) + (1.007825 u)] 
= -0.991 u

Converting to MeV/c2: Q = (-0.991 u) x (931.5 MeV/c2/u) = -920.7 MeV/c2 
Therefore, the correct answer is B) -1.191 MeV (rounded to three decimal places).

Consider the Nuclear Reaction 714N+24He→817O+11H^{14} _{7} N + ^{4}_{2} He → ^{17} _{8} O + ^{1} _{1} H714​N+24​He→817​O+11​H

Consider the Nuclear Reaction $^{14} _{7} N + ^{4}_{2} He → ^{17} _{8} O + ^{1} _{1} H$