Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 465: 34


The element's atomic mass number is 56

Work Step by Step

Let's assume that we have 1 mole of atoms. We can find the number of atoms along each side of the cube. $(6.02\times 10^{23})^{1/3} = 8.44\times 10^7~atoms$ We can find the length $L$ of each side of the cube. $L = (8.44\times 10^7~atoms)(2.27\times 10^{-10}~m)$ $L = 0.01916~m$ We can find the volume of the cube. $V = L^3$ $V = (0.01916~m)^3$ $V = 7.034\times 10^{-6}~m^3$ We can find the mass of the cube. $m = \rho~V$ $m = (7950~kg/m^3)(7.034\times 10^{-6}~m^3)$ $m = 0.0559~kg$ We can find the mass $m_a$ of each atom. $m_a = \frac{m}{6.02\times 10^{23}}$ $m_a = \frac{0.0559~kg}{6.02\times 10^{23}}$ $m_a = 9.236\times 10^{-26}~kg$ We can find the element's atomic mass number. $A = \frac{m_a}{1.66\times 10^{-27}~kg}$ $A = \frac{9.236\times 10^{-26}~kg}{1.66\times 10^{-27}~kg}$ $A = 56$ The element's atomic mass number is 56.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.