Q.
How much energy must be imparted to the proton so as to start the reaction 3Li7+H 1 1 →Be 7 4 +n 1 0 (Given that mass of 3Li7= 7.016 amu ,H 1 1 = 1.00783 amu, Be 7 4 =7.01693 amu and n 1 0 =1.00866 amu,1 amu = 931.5 MeV/C2)
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a
1.27 Mev
b
1.47 Mev
c
1.87 Mev
d
1.67 Mev
answer is C.
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Detailed Solution
Mass of reactants = (7.01600+1.00783)amuMass of products = (7.01693+1.00866)amuTherefore mass defect = (8.02383–8.02559)amu = –0.00176amuThe Q–value of reaction,Q = −0.00176×931.5MeV =−1.64Mev According to conservation law of momentum, the energy supplied in the form of K.E. of proton =−1+mMQ=−1+17−1.64MeV = 1.8742 MeV
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