Chapter 29, Problem 87P

(a)

To determine

Identify the unknown product in the reaction of $\alpha {+}_7^{14}\text{N}\to \text{p}+?$.

Answer to Problem 87P

The unknown product in the reaction of $\alpha {+}_7^{14}\text{N}\to \text{p}+?$ is oxygen and its symbol is ${}_8^{17}\text{O}$.

Explanation of Solution

The given reaction iss

    $\alpha {+}_7^{14}\text{N}\to \text{p}+?$

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, $A:4+14=1+a;a=17$.

Conservation of charge: The number of protons in reactant is equal to the number of protons in the product of the reaction. Therefore, $Z:2+7=1+b;b=8$.

In periodic table, the element with atomic number $28$ is oxygen. Therefore, the reaction product is oxygen and its symbol is ${}_8^{17}\text{O}$

(b)

To determine

The distance between the centers of alpha particle and nitrogen nucleus when they are just touching.

Answer to Problem 87P

The distance between the centers of alpha particle and nitrogen nucleus when they are just touching is $4.8\text{ fm}$.

Explanation of Solution

The mass of number of $\alpha$ particle is $4$ and nitrogen is $14$.

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 $\alpha$ particle is $r=1.2\text{ fm}\times 4^{1/3}=1.9\text{ fm}$

The radius of ${}_7^{14}\text{N}$ nucleus is $r=1.2\text{ fm}\times {14}^{1/3}=2.9\text{ fm}$

Therefore, the distance between the centers of alpha particle and nitrogen nucleus when they are just touching is $4.8\text{ fm}$.

(c)

To determine

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

Answer to Problem 87P

The minimum value of the kinetic energy of the alpha particle and nitrogen nucleus necessary to bring them into contact is $4.2\text{ MeV}$.

Explanation of Solution

The distance between the centers of proton and nitrogen nucleus when they just touching is $4.8\text{ fm}$ and the two point charges are $+2e$ and $+7e$. The charge $e=1.602\times {10}^{-19}\text{ C}$

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.

Substitute $4.8\text{ fm}$ 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 $U_E$

$\begin{array}{l}{U}_E=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 2e\times 7e}{4.8\text{ fm}}\\ U_E=\frac{8.988\times {10}^9\text{ N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\times 14\times 1.602\times {10}^{-19}\text{ C}}{4.8\times {10}^{-15}\text{ m}}\times \frac{1\text{ eV}}{1.602\times {10}^{-19}\text{ J}}\\ U_E=4.2\text{ MeV}\end{array}$

Therefore, the minimum value of the kinetic energy of the alpha particle and nitrogen nucleus necessary to bring them into contact is $4.2\text{ MeV}$.

(d)

To determine

Whether the total kinetic energy of the reaction products more or less than the kinetic energy in part (c). And calculate the kinetic energy difference.

Answer to Problem 87P

The total kinetic energy of the reaction products is less than the initial kinetic energy of the reactants. And the difference in their kinetic energy is $1.1917\text{ MeV}$.

Explanation of Solution

The atomic mass of ${}_2^4\text{He}$ is $4.0026032\text{ u}$, ${}_7^{14}\text{N}$ is $14.0030740\text{ u}$, ${}_1^{1}p$ is $1.0078250\text{ u}$ and ${}_8^{17}\text{O}$ is $16.9991315\text{ u}$.

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

$\begin{array}{l}\Delta m=16.9991315\text{ u}+1.0078250\text{ u}-14.0030740\text{ u}-4.0026032\text{ u}\\ \Delta m=0.0012793\text{ u}\end{array}$

The mass difference is positive, the mass of the product is greater than the mass of the reactant. Therefore the kinetic energy of the product is less than the kinetic energy of the reactant.

Substitute $0.0012793\text{ u}$ for $\Delta m$ and $931.494\text{MeV}/\text{u}$ for $c^2$ in equation (I) to find the value of $E$

$\begin{array}{l}E=0.0012793\text{ u}\times 931.494\text{MeV}/\text{u}\\ E=1.1917\text{ MeV}\end{array}$

Thus, the difference in their kinetic energy is $1.1917\text{ MeV}$.