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: 38

Answer

a) $4.86$ b) $1.458\times 10^9\;\rm K$

Work Step by Step

a) The first reaction of the carbon cycle is given by ​ $$ \rm ^{12}_6C+^2_1H\rightarrow\;\;^{13}_7N+\gamma$$ In example 31-9, we can see that the energy needed is given by $$KE=\dfrac{kQ_{ \rm ^{12}_6C}Q_{\rm ^2_1H}}{r_{ \rm ^{12}_6C}+r_{\rm ^2_1H}}$$ Thus; $$KE=\dfrac{k (6e)(e)}{r_{ \rm ^{12}_6C}+r_{\rm ^2_1H}}$$ $$KE=\dfrac{6k e^2}{r_{ \rm ^{12}_6C}+r_{\rm ^2_1H}}$$ We know that the radius of a nucleus is given by $$r=(1.2\times 10^{-15})A^{\frac{1}{3}}$$ Thus, $$KE=\dfrac{6k e^2}{(1.2\times 10^{-15}) \left[ A_{\rm C}^{\frac{1}{3}}+A_{\rm H}^{\frac{1}{3}} \right]}$$ Plugging the known; $$KE=\dfrac{6(8.99\times 10^9)(1.6\times 10^{-19})^2}{(1.2\times 10^{-15})\left[ 12^{\frac{1}{3}}+1^{\frac{1}{3}} \right]}$$ $$KE=3.498\times 10^{-13}\;\rm\color{red}{\bf\not} J\cdot \dfrac{\rm MeV}{10^6\times 1.6\times 10^{-19}\;\rm\color{red}{\bf\not} J}$$ $$KE=\bf 2.1864\;\rm MeV$$ The ratio needed is given by $$\dfrac{KE_{\rm C}}{KE_{\rm d-t}}=\dfrac{2.1864}{0.45}=\color{red} {\bf 4.86}$$ ------------- b) We know that kinetic energy is proportional to temperature. $$KE\propto T$$ Thus, $$\dfrac{KE_{\rm C}}{KE_{\rm d-t}}=\dfrac{T_{\rm C}}{T_{\rm d-t}}=4.86$$ Solving for $T_{\rm C}$; $$T_{\rm C}=4.86T_{\rm d-t}$$ Plugging the known; $$T_{\rm C}=4.86 \left(3\times 10^8\right)=\color{red} {\bf 1.458\times 10^9}\;\rm K$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.