Chapter 29, Problem 69P

(a)

To determine

The reaction product must be an $\alpha$ particle.

Answer to Problem 69P

The reaction product must be an $\alpha$ particle as the atomic number of the reaction product is 2, its symbol is ${}_2^4\text{He}$.

Explanation of Solution

The given equation of carbon cycle CNO-I that takes place inside stars is

    ${}_1^{1}p{+}_{{}^7}^{15}\text{N}{\to }_6^{12}\text{C}{+}_b^a(?)$

Here, $(?)$ is the unknown reaction product.

Conservation of nucleon number: The number of nucleon number in reactant is equal to the number of nucleon number in the product of the reaction. Therefore, $A:1+15=12+a;a=4$.

Conservation of charge: The number of protons in reactant is equal to the number of protons in the product of the reaction. Therefore, $Z:1+7=6+b;b=2$.

In periodic table, the element with atomic number $2$ is helium. The reaction product must be an $\alpha$ particle as the atomic number of the reaction product is 2, its symbol is ${}_2^4\text{He}$.

(b)

To determine

The amount of energy released by the last step in the carbon cycle CNO-I.

Answer to Problem 69P

The amount of energy released by the last step in the carbon cycle CNO-I is $4.9654\text{ MeV}$.

Explanation of Solution

The atomic mass of ${}_6^{12}\text{C}$ is $12.0000000\text{ u}$, ${}_7^{15}\text{N}$ is $15.0001089\text{ u}$, ${}_1^{1}p$ is $1.0078250\text{ u}$ and ${}_2^4\text{He}$ is $4.0026033\text{ u}$

Write the formula for energy

  $E=\left|\Delta m\right|c^2$                                                           (I)

Here, $E$ is the energy, $\Delta m$ is the mass difference and $c$ is the speed of light $(c^2=931.494\text{MeV}/\text{u})$.

The mass difference between the product and the reactant of the overall equation is

  $\begin{array}{l}\Delta m=12.0000000\text{ u}+4.0026033\text{ u}-15.0001089\text{ u}-1.0078250\text{ u}\\ \Delta m=-0.0053306\text{ u}\end{array}$

Substitute $0.0053306\text{ u}$ for $\Delta m$ and $931.494\text{MeV}/\text{u}$ for $c^2$ in equation (I) to find the value of $E$

  $\begin{array}{l}E=0.0053306\text{ u}\times 931.494\text{MeV}/\text{u}\\ E=4.9654\text{ MeV}\end{array}$

Thus, the amount of energy released by the last step in the carbon cycle CNO-I is $4.9654\text{ MeV}$.

(c)

To determine

The distance between the centers of proton and nitrogen nucleus when they just touching.

Answer to Problem 69P

The distance between the centers of proton and nitrogen nucleus when they just touching is $4.16\text{ fm}$.

Explanation of Solution

Write the equation for radius

    $r=r_0{A}^{1/3}$                                                                              (I)

Here, $r$ is the radius, $r_0$ is the initial radius $r_0=1.2\text{ fm}$ and $A$ is the mass number.

The radius of proton is $r=1.2\text{ fm}\times 1^{1/3}=1.2\text{ fm}$

The radius of ${}_7^{15}\text{N}$ nucleus is $r=1.2\text{ fm}\times {15}^{1/3}=2.96\text{ fm}$

Therefore, the distance between the centers of proton and nitrogen nucleus when they just touching is $4.16\text{ fm}$.

(d)

To determine

The minimum value of the total kinetic energy of the proton and nitrogen nucleus necessary to bring them into contact.

Answer to Problem 69P

The minimum value of the total kinetic energy of the proton and nitrogen nucleus necessary to bring them into contact is $2.4\text{ MeV}$.

Explanation of Solution

Write the formula for electric potential energy

    $U_E=\frac{k{e}^2}{r}$                                                                                       (I)

Here, $U_E$ is the electric potential energy, $k$ is Coulomb constant, $e$ is the charge and $r$ is the radius.

The distance between the centers of proton and nitrogen nucleus when they just touching is $4.16\text{ fm}$ and the two point charges are $+e$ and $+7e$. The charge $e=1.602\times {10}^{-19}\text{ C}$

Substitute $4.16\text{ fm}$ for $r$, $8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ in equation (I) to find the value of $U_E$

  $\begin{array}{c}{U}_E=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times e\times 7e}{4.16\text{ fm}}\\ =\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 7\times 1.602\times {10}^{-19}\text{ C}}{4.16\times {10}^{-15}\text{ m}}\times \frac{1\text{ eV}}{1.602\times {10}^{-19}\text{ J}}\\ =2.4\text{ MeV}\end{array}$

Therefore, the minimum value of the total kinetic energy of the proton and nitrogen nucleus necessary to bring them into contact is $2.4\text{ MeV}$.