Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1151: 4

Answer

N=20

Work Step by Step

(a) We know that $A=\frac{r^3}{(1.2\times 10^{-15})^3}$ We plug in the known values to obtain: $A=(\frac{4.0\times 10^{-15}}{1.2\times 10^{-15}})^3$ $\implies A=37$ Now we can find the number of neutrons as follows: $N=A-Z$ $\implies N=37-17=20$
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