Answer
See answers.
Work Step by Step
a. The reaction is
$$d + \;^{6}_{3}Li\rightarrow\;X +p$$
Conserve nucleon number: 2 + 6 = A + 1. We see that A = 7.
Conserve charge: 1 + 3 = Z + 1. We see that Z = 3.
The unknown product is Lithium-7.
$$d + \;^{6}_{3}Li\rightarrow\; \;^{7}_{3}Li +p$$
b. This is named a "stripping" reaction because deuteron has been stripped of a neutron, by lithium.
c. Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left( m_{d}+m(^{6}_{3}Li)-m(^{7}_{3}Li)-m_{p} \right)c^2$$
$$ =\left(2.014102u+6.015123u-7.016003u-1.007825u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=5.027MeV$$
The Q-value is positive. The reaction is exothermic.
