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

Published by Pearson

# Chapter 36 - Relativity - Exercises and Problems - Page 1062: 71

#### Answer

(a) The amount of energy released in each fusion is $~~4.32\times 10^{-12}~J$ (b) The fraction of the initial rest mass energy that is released is $~~0.00717$

#### Work Step by Step

(a) We can find the mass of four protons: $(4)(1.673\times 10^{-27}~kg) = 6.692\times 10^{-27}~kg$ We can find the difference between the original mass of four protons and the final mass of the helium nucleus: $(6.692\times 10^{-27}~kg)-(6.644\times 10^{-27}~kg) = 4.8\times 10^{-29}~kg$ This "missing mass" after the fusion is the mass that has been converted into energy and released. We can calculate the energy: $E = mc^2$ $E = (4.8\times 10^{-29}~kg)(3.0\times 10^8~m/s)^2$ $E = 4.32\times 10^{-12}~J$ The amount of energy released in each fusion is $~~4.32\times 10^{-12}~J$ (b) The fraction of energy that is released is equal to the fraction of the "missing mass" that is converted to energy compared to the initial mass of four protons. We can find this fraction: $\frac{4.8\times 10^{-29}~kg}{6.692\times 10^{-27}~kg} = 0.00717$ The fraction of the initial rest mass energy that is released is $~~0.00717$

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