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Find the Q value of the reaction. ""_(1)H^(2)+""_(3)Li^(6) to ""_(1)H^(1)+""_(3)Li^(7) The rest mass of ""_(1)H^(2)=2.01410 a.m.u. ""_(3)Li^(6)=6.01513 a.m.u. ""_(3)Li^(7)=7.01601 a.m.u. ""_(1)H^(1)=1.00783 a.m.u. |
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Answer» Solution :`""_(1)H^(2)+""_(3)Li^(6)=""_(3)Li^(6)+""_(1)H^(1)+Q` Then, 8.02933=8.02384+Q Q=0.00549 a.m.u. `=0.00549 XX 931 MeV=5.1MeV` |
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