Chapter 31, Problem 6P

(a)

To determine

Weather the reaction $\mathrm{n}{+}_{12}^{27}\text{Mg}{\to }_{11}^{23}\text{Na}+\text{d}$ occurs if the bombarding particles have 10.00 MeV of the kinetic energy.

Answer to Problem 6P

The energy required is of $-9.469\text{MeV}$ .

Explanation of Solution

Given Information:

The reaction is $\mathrm{n}{+}_{12}^{27}\text{Mg}{\to }_{11}^{23}\text{Na}+\text{d}$ .

Calculation:

Consider the mass defect is,

  $\begin{array}{c}△\mathrm{m}=\left(\text{mass}\text{of}\underset{12}{\overset{24}{\text{Mg}}}+\text{mass}\text{of}\underset{0}{\overset{1}{\mathrm{n}}}\right)-\left(\text{mass}\text{of}\underset{11}{\overset{23}{\mathrm{N}\mathrm{a}}}+\text{mass}\text{of}\underset{1}{\overset{2}{\mathrm{H}}}\right)\\ =\left(23.985042\text{u+1}\text{.008665u}\right)-\left(22.989770\text{u}+\text{2}\text{.014102u}\right)\\ =24.993707\text{u}-\text{25}\text{.000u}\end{array}$

From above it is clear that the mass of the product is more than the mass of the reactants this means that the reaction requires energy.

Consider that if $10\text{MeV}+\text{kinetic}\text{energy}$ is supplied then the energy required is,

  $\begin{array}{c}\mathrm{E}=-\mathrm{\Delta }\mathrm{m}{\mathrm{c}}^{2\mathrm{w}}\\ =\left(-0.010107\times 931.5\right)\text{MeV}\\ =-9.469\text{MeV}\end{array}$

(b)

To determine

The amount of energy released.

Answer to Problem 6P

The energy released is of $0.53\text{MeV}$ .

Explanation of Solution

Calculation:

The energy released from the reaction is,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{E}=10\text{MeV}-9.469\text{MeV}\\ =0.53\text{MeV}\end{array}$