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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