Chapter 30, Problem 4P

To determine

(a) To determine:

The value of A for nucleus having radius $3.7\times {10}^{-15}$ m.

Answer to Problem 4P

Solution:

$R=\text{6}.0\text{25fermi}$

Explanation of Solution

Given:

A=112

Formula used:

$R=r_0{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Calculation:

The radius of ${}_{48}{}^{112}Cd$ is given by,

$\begin{array}{l}R=r_0{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ r_0=\text{constant}=1.25\times {10}^{-15}m\\ A=\text{atomic}\text{mass}\text{number}\\ \text{R=1}\text{.25}\times {10}^{-15}\times 4.820\\ R=6.025\times {10}^{-15}m\\ R=6.025\text{fermi}\end{array}$

Calculation:

The radius of ${}_{48}{}^{112}Cd$ is 6.025fermi.

To determine

(b) To determine:

The value of A for nucleus having radius $3.7\times {10}^{-15}$ m.

Answer to Problem 4P

Solution:

$A=\text{25}.\text{934}$

Explanation of Solution

Given:

$R=3.7\times {10}^{-15}m$

Formula used:

$R=r_0{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Calculation:

The radius of nucleus given by,

$\begin{array}{l}R=r_0{A}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ r_0=\text{constant}=1.25\times {10}^{-15}m\\ A=\text{atomic}\text{mass}\text{number}\\ \text{3}\text{.7}\times {\text{10}}^{-15}\text{=1}\text{.25}\times {10}^{-15}\times \sqrt[3]{A}\\ \sqrt[3]{A}=\frac{\text{3}\text{.7}\times {\text{10}}^{-15}}{\text{1}\text{.25}\times {10}^{-15}}\\ \sqrt[3]{A}=2.96\\ A=25.934\end{array}$

Calculation:

The value of A for nucleus having radius $3.7\times {10}^{-15}$ m is 25.934.