Chapter 30, Problem 43P

(a)

To determine

The radius of the ${}_6^{12}\text{C}$ nucleus.

Answer to Problem 43P

The radius of the ${}_6^{12}\text{C}$ nucleus is $\underset{\_}{2.7\text{fm}}$.

Explanation of Solution

Write the expression for the radius of the ${}_6^{12}\text{C}$ nucleus.

    $r=a{A}^{1/3}$        (I)

Here, $r$ is the radius of the ${}_6^{12}\text{C}$ nucleus, $A$ is the mass number, $a$ is a constant.

Conclusion:

Substitute $1.2\text{fm}$ for $a$, $12$ for $A$ in equation (I) to find $r$.

    $\begin{array}{c}r=\left(1.2\text{fm}\times \frac{{10}^{-15}\text{m}}{1\text{fm}}\right){\left(12\right)}^{1/3}\\ =2.7\times {10}^{-15}\text{m}\\ \text{=2}\text{.7}\text{fm}\end{array}$

Therefore, the radius of the ${}_6^{12}\text{C}$ nucleus is $\underset{\_}{2.7\text{fm}}$.

(b)

To determine

The force of repulsion between a proton at the surface of an ${}_6^{12}\text{C}$ nucleus and the remaining five protons.

Answer to Problem 43P

The force of repulsion between a proton at the surface of an ${}_6^{12}\text{C}$ nucleus and the remaining five protons $\underset{\_}{1.5\times {10}^2\text{N}}$.

Explanation of Solution

Write the expression for the force of repulsion between a proton at the surface of an ${}_6^{12}\text{C}$ nucleus and the remaining five protons.

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

Here, $F$ is the force of repulsion between a proton at the surface of an ${}_6^{12}\text{C}$ nucleus and the remaining five protons, $k_e$ is the columbic constant, $r$ is the radius of the nucleus, $Z$ is the number of protons, $e$ is the electronic charge.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $6$ for $Z$, $2.7\text{fm}$ for $r$, $1.60\times {10}^{-19}\text{C}$ for $e$ in equation (II) to find $F$.

    $\begin{array}{c}F=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(6-1\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{{\left(2.7\text{fm}\times \frac{{10}^{-15}\text{m}}{1\text{fm}}\right)}^2}\\ =1.5\times {10}^2\text{N}\end{array}$

Therefore, the force of repulsion between a proton at the surface of an ${}_6^{12}\text{C}$ nucleus and the remaining five protons $\underset{\_}{1.5\times {10}^2\text{N}}$.

(c)

To determine

The work done to overcome the last proton from a large distance up to the surface of the nucleus.

Answer to Problem 43P

The work done to overcome the last proton from a large distance up to the surface of the nucleus is $\underset{\_}{2.6\text{MeV}}$.

Explanation of Solution

Write the expression for the work done to overcome the last proton from a large distance up to the surface of the nucleus.

    $\begin{array}{c}U=\frac{k_e\left(Z-1\right)e\left(1e\right)}{r}\\ =\frac{k_e\left(Z-1\right)e^2}{r}\end{array}$        (III)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $6$ for $Z$, $2.7\text{fm}$ for $r$, $1.60\times {10}^{-19}\text{C}$ for $e$ in equation (II) to find $U$.

    $\begin{array}{c}U=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(6-1\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{\left(2.7\text{fm}\times \frac{{10}^{-15}\text{m}}{1\text{fm}}\right)}\\ =4.2\times {10}^{-13}\text{J}\\ U\left(\text{in MeV}\right)=\left(4.2\times {10}^{-13}\text{J}\times \frac{1\text{MeV}}{1.60\times {10}^{-13}\text{J}}\right)\\ =2.6\text{MeV}\end{array}$

Therefore, the work done to overcome the last proton from a large distance up to the surface of the nucleus is $\underset{\_}{2.6\text{MeV}}$.

(d)

To determine

The radius of ${}_{92}^{238}\text{U}$ nucleus, the force between a proton at the surface of ${}_{92}^{238}\text{U}$ and the remaining protons, the work done to overcome the last proton from a large distance up to the surface of the nucleus.

Answer to Problem 43P

The radius of ${}_{92}^{238}\text{U}$ nucleus is $\underset{\_}{7.4\text{fm}}$, the force between a proton at the surface of ${}_{92}^{238}\text{U}$ and the remaining protons is $\underset{\_}{3.8\times {10}^2\text{N}}$, the work done to overcome the last proton from a large distance up to the surface of the nucleus is $\underset{\_}{18\text{MeV}}$.

Explanation of Solution

Use equation (I) to solve for the radius of ${}_{92}^{238}\text{U}$ nucleus.

Use equation (II) to solve for force between a proton at the surface of ${}_{92}^{238}\text{U}$ and the remaining protons.

Use equation (III) the work done to overcome the last proton from a large distance up to the surface of the nucleus.

Conclusion:

Substitute $1.2\text{fm}$ for $a$, $238$ for $A$ in equation (I) to find $r$.

    $\begin{array}{c}r=\left(1.2\text{fm}\times \frac{{10}^{-15}\text{m}}{1\text{fm}}\right){\left(238\right)}^{1/3}\\ =7.4\times {10}^{-15}\text{m}\\ \text{=7}\text{.4}\text{fm}\end{array}$

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $92$ for $Z$, $7.4\text{fm}$ for $r$, $1.60\times {10}^{-19}\text{C}$ for $e$ in equation (II) to find $F$.

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

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_e$, $92$ for $Z$, $7.4\text{fm}$ for $r$, $1.60\times {10}^{-19}\text{C}$ for $e$ in equation (II) to find $U$.

    $\begin{array}{c}U=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\left(92-1\right){\left(1.60\times {10}^{-19}\text{C}\right)}^2}{\left(7.4\text{fm}\times \frac{{10}^{-15}\text{m}}{1\text{fm}}\right)}\\ =2.8\times {10}^{-12}\text{J}\\ U\left(\text{in MeV}\right)=\left(2.8\times {10}^{-12}\text{J}\times \frac{1\text{MeV}}{1.60\times {10}^{-13}\text{J}}\right)\\ =18\text{MeV}\end{array}$

Therefore, the radius of ${}_{92}^{238}\text{U}$ nucleus is $\underset{\_}{7.4\text{fm}}$, the force between a proton at the surface of an ${}_{92}^{238}\text{U}$ and the remaining protons is $\underset{\_}{3.8\times {10}^2\text{N}}$, the work done to overcome the last proton from a large distance up to the surface of the nucleus is $\underset{\_}{18\text{MeV}}$.