Chapter 31, Problem 12P

(a)

To determine

The resulting nucleusX .

Answer to Problem 12P

X is ${}_3^7\text{Li}$ .

Explanation of Solution

Given:

The reaction is $d{+}_3^6\text{Li}\to \text{X}+\text{p}$ .

Calculation:

The right side of the reaction has 4 protons and 8 nucleons. The product has 1 proton.

Thus, the X will have 3 protons and 4 neutrons.

Then, ${}_3^{7}X{=}_3^7\text{Li}$

Conclusion:

Therefore, the value of the X is ${}_3^{7}X{=}_3^7\text{Li}$ .

(b)

To determine

To Explain:The reaction is called the stripping reaction.

Explanation of Solution

The complete reaction is:

  $d{+}_3^6\text{Li}{\to }_3^7\text{Li}+\text{p}$

This is called stripping reaction because the part of incident nucleus (deuteron) combines with the target nucleus (lithium) producing a proton.

(c)

To determine

The Q value of the reaction.

Whether the reaction is endothermic or exothermic reaction.

Answer to Problem 12P

  $89.4\text{MeV}$

Explanation of Solution

Given:

  $d{+}_3^6\text{Li}{\to }_3^7\text{Li}+\text{p}$

  $m_d=2.0141\text{u}$

  $\begin{array}{c}{m}_{Li-6}=6.01512\text{u}\end{array}$

  $\begin{array}{c}{m}_{Li-7}=7.01600\text{u}\end{array}$

  $\begin{array}{c}{m}_p=1.0078\text{u}\end{array}$

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 valuecan be calculated as:

  $\begin{array}{c}Q=\left[2.0141+6.0151-7.0160-1.0072\right]\left(931.5\frac{\text{MeV}}{{\text{c}}^{\text{2}}}\right){\text{c}}^{\text{2}}\end{array}$

  $\begin{array}{c}=89.4\text{MeV}\end{array}$

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

Conclusion:

Therefore, the Q value is $89.4\text{MeV}$ .