Chapter 7.4, Problem 56PPS

To determine

The radius of the nucleus of an atom for the table shown below using the formula $r=\left(1.3\times {10}^{-15}\right)\sqrt[3]{A}$ meters, where $\left(A\right)$ is the mass number of the nucleus and $\left(r\right)$ is the radius of the nucleus of an atom.

Atom	Mass Number
Carbon	6
Oxygen	8
Sodium	11
Aluminum	13
Chlorine	17

Answer to Problem 56PPS

  $r_{\text{carbon}}=2.36\times {10}^{-15},r_{\text{oxygen}}=2.6\times {10}^{-15},r_{\text{sodium}}=2.89\times {10}^{-15},r_{\text{aluminum}}=3.06\times {10}^{-15}$ and $r_{\text{chlorine}}=3.34\times {10}^{-15}$

Explanation of Solution

Given information: The formula of the radius of the nucleus of an atom for the table shown below is $r=\left(1.3\times {10}^{-15}\right)\sqrt[3]{A}$ meters, where $\left(A\right)$ representsthe mass number of the nucleus and $\left(r\right)$ is the radius of the nucleus of an atom.

Atom	Mass Number
Carbon	6
Oxygen	8
Sodium	11
Aluminum	13
Chlorine	17

Calculation: Using the formula $r=\left(1.3\times {10}^{-15}\right)\sqrt[3]{A}$ , the radius of nucleus of different atoms of given mass numbers can be evaluated by plugging different mass numbers of given atoms.

  $\begin{array}{c}{r}_{\text{carbon}}=\left(1.3\times {10}^{-15}\right)\sqrt[3]{6}=2.36\times {10}^{-15}\\ r_{\text{oxygen}}=\left(1.3\times {10}^{-15}\right)\sqrt[3]{8}=2.6\times {10}^{-15}\\ r_{\text{sodium}}=\left(1.3\times {10}^{-15}\right)\sqrt[3]{11}=2.89\times {10}^{-15}\\ r_{\text{aluminum}}=\left(1.3\times {10}^{-15}\right)\sqrt[3]{13}=3.06\times {10}^{-15}\\ r_{\text{chlorine}}=\left(1.3\times {10}^{-15}\right)\sqrt[3]{17}=3.34\times {10}^{-15}\end{array}$

Hence, the distance of Mars from the Sun is 141.48 million of mile.