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Consider the following reaction; 11H+613C⟶713N+01nThe atomic masses of the nuclei areM 11H=1.007825u,M 0n1=1.008665u,M 613C=13.00336u,M 713N=13.00574u and 1 amu = 931 MeV. The kinetic energy of proton 11Hrequired to initiate the reaction is

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a

1.99 Mev

b

2.99 MeV

c

3.99 Mev

d

4.99

answer is B.

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Detailed Solution

Required kinetic energy =[m(N)+m(n)−m(H)−m(C)]×931MeV =2.99MeV

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