<p>The sun, like all stars, releases energy through nuclear fusion. In this prob

Question

<p>The sun, like all stars, releases energy through nuclear fusion. In this problem, you will find the total number of fusion reaction events that occur inside the sun every second. You will be considering the proton-proton chain, in which four hydrogen nuclei are converted into a helium nucleus and two positrons. The net reaction for the proton-proton chain is</p>
<p><img src="http://session.masteringphysics.com/render?tex=4%5E%7B1%7D+_%7B1%7D+%7B%5Crm+H%7D+%5Crightarrow+%5C+%5E%7B4%7D+_%7B2%7D%7B%5Crm+He%7D+%2B+2+%5Cbeta+%5E%7B%2B%7D" alt="4^{1} _{1} { m H} ightarrow ^{4} _{2}{ m He} + 2 eta ^{+}" align="middle" /></p>
<p>To find the energy released by this reaction, you will need the following mass data:</p>
<p>mass of&#160;<img src="http://session.masteringphysics.com/render?tex=%5E%7B1%7D+_%7B1%7D+%7B%5Crm+H%7D+%3D+1.007825%5C%3B+%7B%5Crm+u%7D" alt="^{1} _{1} { m H} = 1.007825; { m u}" align="middle" />&#160;and&#160;<img src="http://session.masteringphysics.com/render?tex=%5E%7B4%7D+_%7B2%7D+%7B%5Crm+He%7D+%3D+4.002603%5C%3B+%7B%5Crm+u%7D" alt="^{4} _{2} { m He} = 4.002603; { m u}" align="middle" /></p>
<p>Using the masses of the neutral atoms in your calculation accounts for the energy released by the annihilation of the positrons with electrons, so you can work this problem without reference to the positrons or their rest mass.</p>
<p><strong>PART A:</strong>&#160;The power radiated by the sun is about&#160;<img src="http://session.masteringphysics.com/render?tex=4+%5Ctimes+10%5E%7B26%7D+%5C%3B+%5Crm+W" alt="4 imes 10^{26} ; m W" align="middle" />. Using this number, find the number of proton-proton chain fusion reactions that must go on inside the sun in one second to release enough energy to balance the energy radiated away. (If this condition isn't met, the sun would cool off rapidly.)</p>
<p>&#160;</p>
<p><strong>PART B:</strong>&#160;What mass of hydrogen is converted into helium by the fusion reactions in one second? Use&#160;<img src="http://session.masteringphysics.com/render?tex=1.67+%5Ctimes+10%5E%7B-27%7D%5C%3B+%7B%5Crm+kg%7D" alt="1.67 imes 10^{-27}; { m kg}" align="middle" />&#160;for the mass of one hydrogen nucleus.</p>

Explanation / Answer

?M = 4(M) - M' - 2(Me) Me = electron mass ?M = mass defect and the energy release in single fusion reaction E = ?M*c^2 4.3 *10^-12 J

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