Chapter 31, Problem 8P

(a)

To determine

Whether the reaction ${}_7^{14}\text{N}{\to }_8^{17}\text{O}+\mathrm{p}$ occurs if the incident alpha particle has the kinetic energy of $7.68\text{MeV}$ .

Answer to Problem 8P

The energy is required for the reaction to occur and the energy of $7.68\text{MeV}$ is sufficient for the reaction.

Explanation of Solution

Calculation:

Consider the mass reaction is,

  ${}_7^{14}\text{N}{\to }_8^{17}\text{O}+\mathrm{p}$

Consider the energy of the alpha particle is,

  ${}_2^4\mathrm{H}\mathrm{e}{+}_7^{14}\mathrm{N}{\to }_8^{17}\text{O}{+}_1^1\mathrm{H}$

Consider the mass of,

  $\begin{array}{l}\left(\underset{8}{\overset{17}{\mathrm{O}}}+\underset{1}{\overset{1}{\mathrm{H}}}\right)=16.997131+1.007825\\ {\mathrm{m}}_1=18.004956\text{u}\end{array}$

Consider the mass of,

  $\begin{array}{l}\left(\underset{2}{\overset{4}{\mathrm{H}\mathrm{e}}}+\underset{7}{\overset{14}{\mathrm{N}}}\right)=4.002603+14.003074\\ {\mathrm{m}}_2=18.005677\text{u}\end{array}$

Consider the mass defect is,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{m}=\left({\mathrm{m}}_2-{\mathrm{m}}_1\right)\\ =-0.000721\text{u}\end{array}$

Then, the release in the energy is,

  $\left(-0.000721\text{u}\right)\left(\frac{931.5\mathrm{M}\mathrm{e}\mathrm{V}/{\text{c}}^2}{\mathrm{u}}\right){\mathrm{c}}^2=-0.6716115\text{MeV}$

(b)

To determine

The total kinetic energy of the products.

Answer to Problem 8P

The total energy of the product is $7.0083885\text{MeV}$ .

Explanation of Solution

Calculation:

Thus, the energy of the product is,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{E}=7.68-0.6716115\\ =7.0083885\text{MeV}\end{array}$