Chapter 31, Problem 13P

(a)

To determine

The reason why the reaction is called the pickup reaction.

Explanation of Solution

The reaction is,

  ${}_2^3\text{He}{+}_6^{12}\text{C}{\to }_6^{11}\text{X}{+}_2^4\text{He}$

When a nucleon from target nucleus by the incident nucleus, it is called pick-up reaction.

Here, He-3 picks up a nucleon from C-12 and produces He-4.

Conclusion:

Therefore, the reaction is called the pickup reaction as the neutron from the carbon nucleus is picked up.

(b)

To determine

The resulting nucleus, X.

Answer to Problem 13P

  ${}_6^{11}\text{X}{=}_6^{11}\text{C}$

Explanation of Solution

Given:

The reaction is

  $\begin{array}{l}{}_2^3\text{He}{+}_6^{12}\text{C}\to \text{X}+\alpha \end{array}$

  $\begin{array}{c}{}_2^3\text{He}{+}_6^{12}\text{C}\to \text{X}{+}_2^4\text{He}\end{array}$

Calculation:

The right side of the reaction has 8 protons and 15 nucleons. The product has 2 protons and 4 nucleons.

Thus, X should have 6 protons and 11 neutrons.

  $X{=}_6^{11}\text{C}$

Conclusion:

Therefore, the value of the X is ${}_6^{11}\text{C}$ .

(c)

To determine

The Q value of the reaction.

Whether the reaction is endothermic or exothermic reaction.

Answer to Problem 13P

  $1.86\text{MeV}$

Explanation of Solution

Given:

  ${}_2^3\text{He}{+}_6^{12}\text{C}{\to }_6^{11}\text{X}{+}_2^4\text{He}$

  $m_d=2.0141\text{u}$

  $\begin{array}{c}{m}_{He-3}= 3.0160293\text{ u}\end{array}$

  $\begin{array}{c}{m}_{C-12}=12.00000\text{u}\end{array}$

  $\begin{array}{c}{m}_{C-11}=11.011434\text{u}\end{array}$

  $\begin{array}{c}{m}_{He-4}=4.002602\text{u}\end{array}$

Formula Used:

If the reaction is:

  $\begin{array}{l}A+B\to C\end{array}$

  $\begin{array}{c}Q=\left(m_A+m_B-m_C\right)c^2\end{array}$

Calculation:

The Q factor is calculated as,

  $\begin{array}{c}\text{Q=}\left[ 3.0160293+12.00000-11.011434-4.002602\right]\left(931.5\frac{\text{MeV}}{{\text{c}}^{\text{2}}}\right){\text{c}}^{\text{2}}\end{array}$

  $\begin{array}{c}=1.86\text{MeV}\end{array}$

The value is positive which means that this is an exothermic reaction.

Conclusion:

Therefore, the Q value is $1.86\text{MeV}$ and the reaction is exothermic.