42. (a) We are given the energy release per fusion (calculated in §43-7: Q= 26.7 MeV = 4.28 ×10– 12J) and that four protons are consumed in each fusion event. To find how many sets of four protons are in the sample, we adapt Eq. 42-21: ()2326sam4A1000g6.02 10mol1.5 10.441.0gmolpHMNNM⎛⎞==×=×⎜⎟⎝⎠Multiplying this by Qgives the total energy released: 6.4 ×1014J. It is not required that the answer be in SI units; we could have used MeV throughout (in which case the answer is 4.0 ×1027MeV).
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.