Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 31 - Nuclear Energy; Effects and Uses of Radiation - General Problems - Page 913: 66

Answer

a) $13.9\;\rm MeV$ b) $4.315\;\rm MeV$ c) $3.335\times 10^{10}\;\rm K$

Work Step by Step

a) The energy released is given by $$2\;{\rm \left(^{12}_6C\right)\rightarrow \;^{24}_{12}Mg}+Q$$ where $Q$ is the energy released. $$Q= 2Q_{\rm ^{12}_6C}-Q_{\rm ^{24}_{12}Mg} $$ $$Q=\left(2m_{\rm ^{12}_6C}-m_{\rm ^{24}_{12}Mg}\right)c^2$$ Plugging the known; $$Q=\left(2[12.000000]-23.985042\right)\times 931.5$$ $$Q=\color{red}{\bf 13.9}\;\rm MeV$$ _________________________________________________ b) We can say that the potential energy between the two nuclei will be completely converted to kinetic energy and vice versa. Thus, $$PE_{sys}=KE_{sys}$$ $$PE_{sys}=2KE_{\rm _6^{12}C}$$ Hence, $$KE_{\rm _6^{12}C}=\dfrac{PE_{sys}}{2}=\dfrac{kq_1q_2}{2r}$$ $$KE_{\rm _6^{12}C} =\dfrac{k(6e)(6e)}{2r}$$ $$KE_{\rm _6^{12}C} =\dfrac{k(6e)^2}{2r}$$ Plugging the known; $$KE_{\rm _6^{12}C} =\dfrac{8.99\times 10^9(6\times 1.6\times 10^{-19})^2}{2(6\times 10^{-15})}\times \dfrac{1}{10^6\times 1.6\times 10^{-19}}$$ the last term is to convert from J to MeV. $$KE_{\rm _6^{12}C} =\color{red}{\bf 4.315}\;\rm MeV$$ _________________________________________________ c) We know that $$KE=\frac{3}{2}kT$$ Solving for $T$; $$T=\dfrac{2KE}{3k}$$ Plugging from above and recalling to convert the kinetic energy into joules. $$T=\dfrac{2\times 4.315\times 10^6\times 1.6\times 10^{-19}}{3\times 1.38\times 10^{-23}}$$ $$T=\color{red}{\bf3.335\times 10^{10}}\;\rm K$$
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