Chapter 31, Problem 7P

(a)

To determine

Whether the reaction $\mathrm{p}{+}_3^7\text{Li}{\to }_2^4\text{He}+\mathrm{\alpha }$ occurs if the incident protonhas the kinetic energyof $2500\text{KeV}$ .

Answer to Problem 7P

The release in the energy is of $17.35\text{MeV}$ .

Explanation of Solution

Calculation:

Consider the mass reaction is,

  $\mathrm{p}{+}_3^7\text{Li}{\to }_2^4\text{He}+\mathrm{\alpha }$

Consider the mass of $0\left(2\times \underset{2}{\overset{4}{\mathrm{H}\mathrm{e}}}\right)$ is $2\times 4.002603\text{u}$ . Then, the mass is ${\mathrm{m}}_1=8.005206\text{u}$ .

Consider the mass,

  $\begin{array}{c}\left(\underset{\text{1}}{\overset{\text{1}}{\text{H}}}\text{+}\underset{\text{3}}{\overset{\text{7}}{\text{Li}}}\right)=\left(1.007825+7.01600\right)\mathrm{u}\\ {\mathrm{m}}_2=8.023829\text{u}\end{array}$

Consider the mass defect is,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{m}=\left({\mathrm{m}}_2-{\mathrm{m}}_1\right)\\ =0.018623\text{u}\end{array}$

Then, the release in the energy is,

  $\left(0.018623\text{u}\right)\left(\frac{931.5\text{MeV}/{\text{c}}^{\text{2}}}{\mathrm{u}}\right){\mathrm{c}}^2=17.35\text{MeV}$

(b)

To determine

The total kinetic energy of the products.

Answer to Problem 7P

The total energy of the product is $+19.85\text{MeV}$ .

Explanation of Solution

Calculation:

The energy of the incident proton is,

  $\begin{array}{c}\underset{\text{1}}{\overset{\text{1}}{\text{H}}}=2500\text{keV}\\ =2.5\text{MeV}\end{array}$

Thus, the energy of the product is,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{E}=\mathrm{K}{\mathrm{E}}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}}+\mathrm{Q}\\ =\left(17.347+2.5\text{V}\right)\text{MeV}\\ =+19.85\text{MeV}\end{array}$