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Consider the nuclear reaction 2_1H+14_7N-->X+10_5B where X is a nuclide.

a) Calculate the reaction energy Q (in MeV).

X should be 6_3Li?

b) If the 21H nucleus is incident on a stationary 147N nucleus, what minimum kinetic energy must it have for the reaction to occur?

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The Z,A =(3, 6 ) for reactant

Amount of energy relesed for this reaction will be

Sum of the masses of reactants = 14.0067 a.m.u(N)+ 1.0079 a.m.u(H) = 15.0146
Sum of the masses of products =10.811(B) amu + 6.941(X) amu = 17.752
Therefore, ?M = Sum of the masses of products � Sum of the masses of reactants

= 17.752 � 15.0146 = 2.7374a.m.u.

As there is increase of mass, hence energy is being absorbed .
The value of Q will be therefore negative i.e.
Q = 2.7374 a.m.u.X 931.5 MeV/a.m.u = 2549.8881 = 2550 MeV
Hence the above reaction is endogenic .

To conserve both momentum and energy, incoming particles must have a minimum amount of kinetic energy, called the threshold energy
m is the mass of the incoming particle
M is the mass of the target particle
If the energy is less than this amount, the reaction cannot occur

hence ,

KE min = (1+ 1.0079 a.m.u/14.0067 a.m.u(N))|Q|

KE min = 2733.4939 MeV
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