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 - Problems - Page 912: 30

Answer

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Work Step by Step

For the first reaction, which we need to rewrite as $\left[\rm ^1_1H+^1_1H\rightarrow \;\;^2_1H+e^++e^-+\nu\right]$ to balance the charge, is given by $$ Q=Q_{\rm 2(^1_1H)}-Q_{^2_1H}-Q_{2e}$$ $$ Q=\left[2m_{\rm 2(^1_1H)}-m_{^2_1H}-m_{2e}\right]c^2$$ And to find the energy in MeV, we need to multiply the right side by $931.5\dfrac{\rm MeV}{c^2}$ $$ Q=\left[2m_{\rm 2(^1_1H)}-m_{^2_1H}-m_{2e}\right]c^2\left(931.5\dfrac{\rm MeV}{c^2}\right)$$ Plugging from appendix B; $$ Q=\left[2(1.007825)-(2.014102)-2\left(\dfrac{9.11\times 10^{-31}}{1.67\times 10^{-27}}\right)\right]c^2\left(931.5\dfrac{\rm MeV}{c^2}\right)$$ $$Q=\bf 0.42567\approx\bf 0.42\;\rm MeV$$ which is the same energy mentioned in equation 31-6a. ___________________________________ For the second reaction, $\left[\rm ^1_1H+^2_1H\rightarrow \;\;^3_2He+\gamma\right]$, is given by $$ Q=Q_{\rm (^1_1H)}+Q_{\rm (^2_1H)}-Q_{^3_2He} $$ $$ Q=\left[ m_{\rm (^1_1H)}+m_{\rm (^2_1H)}-m_{^3_2He} \right]c^2$$ $$ Q=\left[ m_{\rm (^1_1H)}+m_{\rm (^2_1H)}-m_{^3_2He} \right]c^2\left(931.5\dfrac{\rm MeV}{c^2}\right)$$ Plugging from appendix B; $$ Q=\left[ 1.007825+ 2.014102-3.016029 \right]c^2\left(931.5\dfrac{\rm MeV}{c^2}\right)$$ $$Q=\bf 5.49399\approx\bf 5.49\;\rm MeV$$ which is the same energy mentioned in equation 31-6b. ___________________________________ For the third reaction, $\left[\rm ^3_2He+^3_2He\rightarrow \;\;^4_2He+^1_1H+^1_1H\right]$, is given by $$ Q=2Q_{\rm (^3_2He)}-Q_{\rm (^4_2He)}-2Q_{\rm ^1_1H } $$ $$ Q=\left[ 2 m_{\rm (^3_2He)}-m_{\rm (^4_2He)}-2m_{\rm ^1_1H } \right]c^2$$ $$ Q=\left[ 2 m_{\rm (^3_2He)}-m_{\rm (^4_2He)}-2m_{\rm ^1_1H } \right]c^2\left(931.5\dfrac{\rm MeV}{c^2}\right)$$ Plugging from appendix B; $$ Q=\left[ 2(3.016029)- (4.002603)-2(1.007825)\right]c^2\left(931.5\dfrac{\rm MeV}{c^2}\right)$$ $$Q=\bf 12.8594\approx\bf 12.86\;\rm MeV$$ which is the same energy mentioned in equation 31-6c.
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