( )5pool3227(10/18) 4.32 10 kg(10 / 18)1.44 10 .1.67 10kgpMNm−×===××Using Eq. 42-20, we obtain RNT×≈ln.ln./2144 102101123232chydecay y 4. Since the density of water is ρ= 1000 kg/m3= 1 kg/L, then the total mass of the pool is V= 4.32 ×105kg, where Vis the given volume. Now, the fraction of that mass made up by the protons is 10/18 (by counting the protons versus total nucleons in a water
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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.