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: 64

Answer

$2.01\times 10^{20}\;\nu$

Work Step by Step

We need to find how many neutrinos are being generated and we can see that, for the net proton cycle, $$\rm 4\left(^1_1H\right)\rightarrow\;\;^4_2He+2e^++2\nu+2\gamma$$ that there are 2 neutrinos produced for every 4 protons consumed. This means that the number of neutrinos that comes out of the sun per second is half the number of protons consumed. Thus, we need to find the number of protons that we can find in the solution of problem 62 as the author told us. Therefore, $$\dfrac{N_{\nu}}{t}=\dfrac{N_{e^+}}{2t}=\dfrac{3.489\times 10^{38}}{2\times 1}=\bf 1.7445\times 10^{38}\;\rm \nu/s$$ Let's assume that the neutrinos are uniformly distributed (spherically) when they took off from the sun. So that the number of neutrinos that pass through the area of 180 m$^2$ in one hour is given by $$N_\nu = 1.7445\times 10^{38}\times \dfrac{180 }{4\pi R^2}\cos 60^\circ \times 60^2$$ where $R$ is the distance between our ceiling and the sun, $40^\circ $ is the angle between the neutrinos flux and the ceiling, and $60^2$ is the second in one hour. Thus; $$N_\nu = 1.7445\times 10^{38}\times \dfrac{180 }{4\pi \left[1.4960\times 10^{11}\right]^2}\cos 60^\circ \times 60^2$$ $$N_\nu=\color{red}{\bf 2.01\times 10^{20}}\;\nu$$
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