Question 30.8: An endoergic reaction Now let’s calculate the reaction energ...

SET UP As in Example 30.7, we use the rest masses found in Table 30.2 to evaluate the reaction energy Q, given by Equation 30.12. The initial and total masses each include nine electron masses.

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

SOLVE The mass calculation, in tabular form, is

\mathrm{^4_2He  +  ^{14}_7H  \rightarrow  ^{17}_8O  +  ^1_1H.}                    (30.11)

\mathrm{\begin{matrix} A:  ^4_2He & 4.002603  u & C:^{17}_8O & 16.999132  u \end{matrix} }

\mathrm{\begin{matrix} B:  ^{14}_7N&\frac{14.003074  u}{18.005677  u} &     D:  ^1_1H & \frac{1.007825  u}{18.006957  u} \end{matrix} }

We see that the total rest mass increases by 0.001280 u, and the corresponding reaction energy is

Q = (-0.001280 u)(931.5 MeV/u) = -1.192 MeV.

REFLECT This amount of energy is absorbed in the reaction. In a head-on collision with zero total momentum, the minimum total initial kinetic energy for this reaction to occur is 1.192 MeV. Ordinarily, though, this reaction would be produced by bombarding stationary ^{14}N nuclei with α particles. In this case, the α energy must be greater than 1.192 MeV. The α can’t give up all of its kinetic energy because then the final total kinetic energy would be zero and momentum would not be conserved. It turns out that, to conserve momentum, the initial a energy must be at least 1.533 MeV.

Practice Problem: Consider the reaction \mathrm{^6_3Li  +  ^4_2He \rightarrow  ^9_4Be  +  ^1_1H,} produced by bombarding a solid lithium target with α particles. Show that this reaction is endoergic, and find the amount by which the total initial kinetic energy exceeds the total final value. Answer: 2.125 MeV.