Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 18 - A Macroscopic Description of Matter - Exercises and Problems - Page 511: 8


The element's atomic mass number is 24 amu.

Work Step by Step

We can find the mass of one atom as: $m_a = \frac{\rho}{\rho_N}$ $m_a = \frac{1750~kg/m^3}{4.39\times 10^{28}~atoms/m^3}$ $m_a = 3.986\times 10^{-26}~kg$ We can find the total number of protons or neutrons in the atom as: $A = \frac{m_a}{1.66\times 10^{-27}~kg}$ $A = \frac{3.986\times 10^{-26}~kg}{1.66\times 10^{-27}~kg}$ $A = 24~amu$ The element's atomic mass number is 24 amu.
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