problem 11 hint: find q value, kinematic threshold energy and minimum kinetic en

Question

problem 11 hint: find q value, kinematic threshold energy and minimum kinetic energy of products.

Be Be oth expression for Q (i e, as a function of mass and of the binding energy, BE In of the interactions in question 2, approximate how much work each is done by the projectile to overcome the Coulomb force, assuming the distance of the closest approach is approximately equal to the nuclear radius. 10. Consider the nuclear reaction 10 m B n 9 Be p Be 3H Li a What is the necessary condition for each exit channe 11. The nuclear reaction n 6Li forms the compound nucleus TLi which may decay by one of these five routes 7Li 6Li t n. He p 5He da What is the necessary condition for each exit channel? 12. For parts (a) (d) in question 2, determine the minimum value of the kinetic energy of the projectile necessary to cause the interaction. 16 18.

Explanation / Answer

Mass of n = 1.0888665 u

Li-6    = 6.015122 u

Li-7     = 7.016004 u

He -6 = 6.018889 u

He-5 = 5.012225 u

He-4 (alpha) = 4.002603 u

D(H-2) = 2.014102 u

P(H-1) = 1.007825 u

consider each of the reactions

Li-6 + n -> Li-7 + gama

Q = (6.015122 + 1.008665) - 7.016004   = 0.007783 u

                   = 7.25 Mev

It is exho thermic reaction and can take place at any energy of the neutron , no thershold energy is required

-> Li-6 + n

It is elastic scattering and no compund nucleaus may be formd and does not nedd any thershold energy

-> He-6 + D

Q = (6.015122 + 1.008665) - (6.018889 + 1.007825)

    = -0.002927 u   = -2.73 Mev

It is an endothermic reaction and the input neutron must have a thershold KE in addition to Q to conserve Momentum

Kth = -Q(1+ Mn/MLi6   = 2.73( 1+ 1.008665/6.015122) = 3.19 Mev

-> He-5 + D

Q = (6.015122 + 1.008665) - (5.012225 + 2.014102) = -0.002540 u

         = -2.37 Mev

It is an endothermic reacrtion and the neutron must have a thershold KE in addition to Q to conserve momentum

Kth = -Q(1+ Mn/MLi6   = 2.37( 1+ 1.008665/6.015122) = 2.77 Mev

-> H-3 + He-4

Q = (6.015122 + 1.008665) - (3.016049+4.002603) = 0.005135 u

       = 4.78 Mev

It is an exhothermic reaction and does not need any thershold energy for the neutron.

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