Question 30.7: An exoergic reaction Now let’s use Equation 30.12 to calcula...

SET UP The reaction can be represented as

\mathrm{^1_1 H  +  ^7_3Li  \rightarrow {}^4_2He  +  ^4_2He.}

We need to use the definition of reaction energy, Equation 30.12. The masses we need are given in Table 30.2.

SOLVE From Table 30.2, we find the initial and final masses:

\mathrm{\begin{matrix} A: {}^1_1H & 1.007825  u & C:^4_2He& 4.002603  u \end{matrix} }

\mathrm{\begin{matrix} B: {^7_3Li} &\frac{7.016005  u}{8.023830  u} &     D: {}^4_2He &\frac{4.002603  u}{8.005206  u} \end{matrix} }

Four electron masses are included on each side. We see that

\mathrm{M_A  +  M_B  –  M_C  –  M_D = 0.018624  u.}

The mass decreases by 0.018624 u; from Equation 30.12, the reaction energy is

\mathrm{Q = (M_A+M_B-M_C-M_D)}c^2.                                (30.12)

Q = (0.018624 u)(931.5 MeV/u) = 17.35 MeV.

REFLECT The total mass decreases, so the total kinetic energy increases. The final total kinetic energy of the two separating α particles is 17.35 MeV greater than the initial total kinetic energy of the proton and the lithium nucleus.

Practice Problem: Find the Q value for the reaction \mathrm{^1_0n  +  ^{10}_5B  \rightarrow  ^7_3Li  +  ^4_2He.} Is this reaction exoergic or endoergic? Answers: 2.79 MeV, exoergic.