Chapter 30, Problem 18E

(a)

To determine

The nuclei radii for ${}^4{\text{He, }}^{27}{\text{Al, }}^{64}{\text{Cu, }}^{125}\text{I}$, and ${}^{216}\text{Po}$.

Answer to Problem 18E

The nuclei radii for ${}^4{\text{He, }}^{27}{\text{Al, }}^{64}{\text{Cu, }}^{125}\text{I}$, and ${}^{216}\text{Po}$ is $2.2\text{fm, }4.2\text{fm},5.6\text{fm},7.0\text{fm}$, and $8.4\text{fm}$ respectively.

Explanation of Solution

Given that the mass number for ${}^4{\text{He, }}^{27}{\text{Al, }}^{64}{\text{Cu, }}^{125}\text{I}$, and ${}^{216}\text{Po}$ is $4,27,64,125$ and $216$ respectively.

The expression for the radius of the nucleus is,

$R=R_o{A}^{1/3}$        (I)

Here, $R$ is the radius, $R_o$ is the absolute radius which is $1.4\times {10}^{-15}\text{m}$, and $A$ is the mass number.

Conclusion:

For ${}^4\text{He}$,

Substitute $1.4\times {10}^{-15}\text{m}$ for $R_o$ and $4$ for $A$ in Equation (I) to find the nuclei radii for ${}^4\text{He}$.

  $\begin{array}{c}R=\left(1.4\times {10}^{-15}\text{m}\right){\left(4\right)}^{1/3}\\ =2.2\times {10}^{-15}\text{m}\left(\frac{1\text{fm}}{{10}^{-15}\text{m}}\right)\\ =2.2\text{fm}\end{array}$

For ${}^{27}\text{Al}$,

Substitute $1.4\times {10}^{-15}\text{m}$ for $R_o$ and $27$ for $A$ in Equation (I) to find the nuclei radii for ${}^{27}\text{Al}$.

  $\begin{array}{c}R=\left(1.4\times {10}^{-15}\text{m}\right){\left(27\right)}^{1/3}\\ =4.2\times {10}^{-15}\text{m}\left(\frac{1\text{fm}}{{10}^{-15}\text{m}}\right)\\ =4.2\text{fm}\end{array}$

For ${}^{64}\text{Cu}$,

Substitute $1.4\times {10}^{-15}\text{m}$ for $R_o$ and $64$ for $A$ in Equation (I) to find the nuclei radii for ${}^{64}\text{Cu}$.

  $\begin{array}{c}R=\left(1.4\times {10}^{-15}\text{m}\right){\left(64\right)}^{1/3}\\ =5.6\times {10}^{-15}\text{m}\left(\frac{1\text{fm}}{{10}^{-15}\text{m}}\right)\\ =5.6\text{fm}\end{array}$

For ${}^{125}\text{I}$,

Substitute $1.4\times {10}^{-15}\text{m}$ for $R_o$ and $125$ for $A$ in Equation (I) to find the nuclei radii for ${}^{125}\text{I}$.

  $\begin{array}{c}R=\left(1.4\times {10}^{-15}\text{m}\right){\left(125\right)}^{1/3}\\ =7.0\times {10}^{-15}\text{m}\left(\frac{1\text{fm}}{{10}^{-15}\text{m}}\right)\\ =7.0\text{fm}\end{array}$

For ${}^{216}\text{Po}$,

Substitute $1.4\times {10}^{-15}\text{m}$ for $R_o$ and $216$ for $A$ in Equation (I) to find the nuclei radii for ${}^{216}\text{Po}$.

  $\begin{array}{c}R=\left(1.4\times {10}^{-15}\text{m}\right){\left(216\right)}^{1/3}\\ =8.4\times {10}^{-15}\text{m}\left(\frac{1\text{fm}}{{10}^{-15}\text{m}}\right)\\ =8.4\text{fm}\end{array}$

Therefore, the nuclei radii for ${}^4{\text{He, }}^{27}{\text{Al, }}^{64}{\text{Cu, }}^{125}\text{I}$, and ${}^{216}\text{Po}$ is $2.2\text{fm, }4.2\text{fm},5.6\text{fm},7.0\text{fm}$, and $8.4\text{fm}$ respectively.

(b)

To determine

To draw the graph for the nuclear radius versus mass number.

Answer to Problem 18E

The graph for the nuclear radius versus mass number is,

Explanation of Solution

From part (a) the nuclei radii for ${}^4{\text{He, }}^{27}{\text{Al, }}^{64}{\text{Cu, }}^{125}\text{I}$, and ${}^{216}\text{Po}$ is $2.2\text{fm, }4.2\text{fm},5.6\text{fm},7.0\text{fm}$, and $8.4\text{fm}$ respectively.

The below figure (1) shows the graph for the nuclear radius versus mass number,

This graph shows a straight line except for the lighter nuclei. This indicates that the volume is proportional to the mass of the nuclei. And the density is almost constant for the nuclei.

Conclusion:

The graph for the nuclear radius versus mass number is,