Chapter 44, Problem 57AP

(a)

To determine

What is the radius of the ${}_6^{12}\text{C}$ nucleus?

Answer to Problem 57AP

The radius of the ${}_6^{12}\text{C}$ nucleus is $\text{2}\text{.7 fm}$.

Explanation of Solution

Write the equation for radius.

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

Here, $r$ is the radius of the nucleus, $a$ is the constant and $A$ is the mass number.

Conclusion:

Substitute $1.2\times {10}^{-15}\text{ m}$ for $a$ and $12$ for $A$ in equation I.

    $\begin{array}{c}r=\left(1.2\times {10}^{-15}\text{ m}\right){\left(12\right)}^{1/3}\\ =2.7\text{ fm}\end{array}$

Therefore, the radius of the ${}_6^{12}\text{C}$ nucleus is $\text{2}\text{.7 fm}$.

(b)

To determine

What is the force of repulsion between a protons on the surface to the remaining five proton?

Answer to Problem 57AP

The force of repulsion between a protons on the surface to the remaining five proton is $\text{1}\text{.5}\times {\text{10}}^2\text{ N}$.

Explanation of Solution

Write the equation for force of repulsion.

    $F=\frac{k_e\left(Z-1\right)e^2}{r^2}$                                                    (II)

Here, $r$ is the radius of the nucleus, $F$ is the force of repulsion, $k_e$ is the constant, $Z$ is the atomic number and $e$ is the charge of electron.

Conclusion:

Substitute $\text{2}\text{.7 fm}$ for $r$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_e$, $1.60\times {10}^{-19}\text{ C}$ for $e$ and $6$ for $Z$ in equation II.

    $\begin{array}{c}F=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(6-1\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{{\left(2.7\times {10}^{-15}\text{ m}\right)}^2}\\ =\text{1}\text{.5}\times {\text{10}}^2\text{ N}\end{array}$

Therefore, the force of repulsion between a protons on the surface to the remaining five proton is $\text{1}\text{.5}\times {\text{10}}^2\text{ N}$.

(c)

To determine

Find the work has to be done to overcome the repulsion force in transporting the last proton from large distance to the surface of nucleus.

Answer to Problem 57AP

The work has to be done to overcome the repulsion force in transporting the last proton from large distance to the surface of nucleus is $\text{2}\text{.6 MeV}$.

Explanation of Solution

Write the equation for work.

    $U=\frac{k_e{q}_1{q}_2}{r}$                                                    (III)

Here, $r$ is the radius of the nucleus, $U$ is the work, $k_e$ is the constant, $q_1$ and $q_2$ are charge.

Conclusion:

Substitute $\text{2}\text{.7 fm}$ for $r$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_e$, $1.60\times {10}^{-19}\text{ C}$ for $q_1$ and $q_2$ in equation III.

    $\begin{array}{c}U=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{\left(2.7\times {10}^{-15}\text{ m}\right)}\\ =2.6\text{ MeV}\end{array}$

Therefore, the work has to be done to overcome the repulsion force in transporting the last proton from large distance to the surface of nucleus is $\text{2}\text{.6 MeV}$.

(d)

To determine

Find the radius of nucleus, repulsion force and work has to be done to overcome the repulsion force in transporting the last proton from large distance to the surface of nucleus in ${}_{92}^{238}\text{U}$.

Answer to Problem 57AP

The radius of nucleus, repulsion force and work has to be done to overcome the repulsion force in transporting the last proton from large distance to the surface of nucleus in ${}_{92}^{238}\text{U}$ are $7.4\text{ fm}$, $3.8\times {\text{10}}^2\text{ N}$ and $\text{18 MeV}$ respectively.

Explanation of Solution

Use the previous part equations.

Conclusion:

Substitute $1.2\times {10}^{-15}\text{ m}$ for $a$ and $238$ for $A$ in equation I.

    $\begin{array}{c}r=\left(1.2\times {10}^{-15}\text{ m}\right){\left(238\right)}^{1/3}\\ =7.4\text{ fm}\end{array}$

Substitute $\text{7}\text{.4 fm}$ for $r$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_e$, $1.60\times {10}^{-19}\text{ C}$ for $e$ and $92$ for $Z$ in equation II.

    $\begin{array}{c}F=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(92-1\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{{\left(7.4\times {10}^{-15}\text{ m}\right)}^2}\\ =3.8\times {\text{10}}^2\text{ N}\end{array}$

Substitute $\text{7}\text{.4 fm}$ for $r$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_e$, $1.60\times {10}^{-19}\text{ C}$ for $q_1$ and $q_2$ in equation III.

    $\begin{array}{c}U=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left(1.60\times {10}^{-19}\text{ C}\right)}^2}{\left(7.4\times {10}^{-15}\text{ m}\right)}\\ =18\text{ MeV}\end{array}$

Therefore, the radius of nucleus, repulsion force and work has to be done to overcome the repulsion force in transporting the last proton from large distance to the surface of nucleus in ${}_{92}^{238}\text{U}$ are $7.4\text{ fm}$, $3.8\times {\text{10}}^2\text{ N}$ and $\text{18 MeV}$ respectively.