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: 61

Answer

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Work Step by Step

a) The author asks us to write the decay equation that gives the radon gas $\rm ^{222}_{86}Rn$ and $\rm ^4_2He$ $$\rm ^{226}_{88}Ra\rightarrow\;\;^{222}_{86}Rn+\;^4_2He$$ ___________________________________________ b) Since we will ignore the kinetic energy of the daughter nucleus, the kinetic energy of $\alpha$ is given by $$KE_{\rm ^4_2He}=Q_{\rm ^{226}_{88}Ra}-Q_{\rm ^{222}_{86}Rn}-Q_{\rm ^4_2He}$$ $$KE_{\rm ^4_2He}=\left(m_{\rm ^{226}_{88}Ra}-m_{\rm ^{222}_{86}Rn}-m_{\rm ^4_2He}\right)931.5$$ Plugging the known; $$KE_{\rm ^4_2He}=\left( 226.025410- 222.017578-4.002603\right) \times 931.5$$ $$KE_{\rm ^4_2He}=\color{red}{\bf 4.871}\;\rm MeV$$ ___________________________________________ c) Since the energy of the alpha particle is less than the threshold of relativism, we can use the momentum law. And since the momentum is conserved, then $$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}$$ where $p=\sqrt{2mKE}$; Thus, $$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}=\sqrt{2m_{\rm ^4_2He}KE_{\rm ^4_2He}}$$ Plugging the known; $$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}=\sqrt{2\times 4.002603 \times 4.871\times 931.5 }$$ $$p_{\rm ^4_2He}=p_{\rm ^{222}_{86}Rn}=\color{red}{\bf 190.6}\;{\rm MeV}/c$$ ___________________________________________ d) We know that $$KE=\dfrac{p^2}{2m}$$ Thus, $$KE_{\rm ^{222}_{86}Rn}=\dfrac{p_{\rm ^{222}_{86}Rn}^2}{2m_{\rm ^{222}_{86}Rn}}$$ Plugging the known; $$KE_{\rm ^{222}_{86}Rn}=\dfrac{190.6^2}{2(222.017578)(931.5)}$$ $$KE_{\rm ^{222}_{86}Rn}= \color{red}{\bf 0.0878}\;\rm MeV$$ Now it is obvious that it was valid to ignore the kinetic energy of the daughter nuclei since its energy is less than $2\%$ of the entry of particle alpha. $$\dfrac{KE_{_{\rm ^{222}_{86}Rn}}}{KE_{\rm ^4_2He}}=\dfrac{0.0878}{4.871}$$ $${KE_{_{\rm ^{222}_{86}Rn}}}=\bf0.018\;{KE_{\rm ^4_2He}}$$
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