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 (reaction energy) can be calculated using the formula:

Q = (m_initial - m_final) c^2

Where m_initial and m_final are the initial and final masses of the nuclei involved in the reaction, and c is the speed of light.

Substituting the given atomic masses, we get:

Q = [(14.003074 u + 4.002603 u) - (16.999131 u + 1.007825 u)] x 931.5 MeV/c^2/u

Q = (-0.992279 u) x 931.5 MeV/c^2/u

Q = -923.1 MeV

Therefore, the Q-value of this reaction is -923.1 MeV.

However, the question asks for the Q-value in MeV/c^2, which is obtained by dividing by the speed of light squared (c^2):

Q/c^2 = -923.1 MeV / (299792458 m/s)^2

Q/c^2 = -1.1908 MeV

Rounding to three significant figures, we get:

Q/c^2 = -1.191 MeV

Therefore, the best choice is A, with an explanation as shown.

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