()( )31421HHHe3.01605u+1.00783u4.00260u 931.5MeV/u19.8MeV.Emmmc∆=+ −=−=(b) The second step is to add energy to produce 3HH.→+n2The energy needed is ( )23222.01410u+1.00867u 3.01605u 931.5MeV/u6.26MeV.nEmmmc=−=(c) The third step: 2H→+pn, which — to make the electrons “balance” — may be rewritten as 2+→1n.The work required is ( )( )12231.00783u 1.00867u2.01410u 931.5MeV/u2.23MeV.nmc=+−=(d) The total binding energy is be123EEEE∆+∆+∆=19.8MeV 6.26MeV
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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.