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Asked by swayamagarwal2114 | 21 Jul, 2022, 05:17: PM
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If the energy of incident deutron  is 10 MeV, then we get speed u of deutron from the following expression
 
Ek = (1/2) m u2
 
where m is mass of deutron
 
begin mathsize 14px style u space equals space square root of fraction numerator 2 space E over denominator m end fraction end root space equals space square root of fraction numerator 2 cross times 10 cross times 1.602 space cross times 10 to the power of negative 13 end exponent over denominator 2.01472 space cross times space 1.66 space cross times space 10 to the power of negative 27 end exponent end fraction end root space equals space 3.095 space cross times space 10 to the power of 7 space m divided by s end style
As shown in figure , let us assume deutron incident along x-axis direction with velocity u on a
Li nucleus which is at rest . The collision makes a following nuclear reaction.
 
begin mathsize 14px style D presubscript 1 presuperscript 2 space plus space L presubscript 3 presuperscript 7 i space rightwards arrow B presubscript 4 presuperscript 8 e plus n presubscript 0 presuperscript 1 end style
 
 
hence , at end of nuclear reaction, we get Be and neutron. Neutron is moving with velocity v along y-axis.
Let Be moves with velocity V in a direction that makes angle θ with x-axis as shown in figure.
 
Total mass before collision = ( 2.01472 + 7.01784 ) amu = 9.03256 amu
Total mass after  collision  = ( 1.00893 + 8.00776 ) amu = 9.01669 amu
 
Tha mass difference Δm is converted into energy of nuclear reaction , E = Δm c2 , where c is speed of light 
 
Δm = 1.587 × 10-2 amu   ;   E = ( 1.587 × 10-2 × 931.5 ) MeV = 14.783 MeV = 2.368 × 10-12 J
 
( Conversion factors used abve :-   1 amu = 931.5 MeV , 1 MeV = 1.602 × 10-13 J )
 
Collision is super elastic collision because energy is increased after collision .
 
By momenum conservation ,  md u = mBe Vx
 
where md is mass of deutron , mBe is mass of Be nucleus and Vx is x-component of velocity of Be nucleus
 
Vx = ( md / mBe ) u  = ( 2.01472 / 8.00776 ) × 3.095 × 107  m/s
 
Vx = 7.787 × 106 m/s   .................................(1)

By momenum conservation ,  mn v = mLi Vy
 
Hence speed of neutron v = ( mLi / mn ) Vy  = ( 7.01784 / 1.00893 ) Vy
 
speed of neutron v = 6.9557 Vy ..............................(2)
 
If energy released from nuclear reaction appears in the form of kinetie energies of neutron and Be nucleus,
then we have
 
(1/2) mn v2 + (1/2) mBe Vx2 + (1/2) mBe Vy2  = 2.368 × 10-12 J
 
By substituting Vx from eqn.(1) and substituting v from eqn.(2) ,
we get from above expression ,  Vy = 6.455 × 106 m/s
 
from eqn.(2) , speed of neutron v = 4.515 × 107 m/s
 
Velocity of Be nucleus = begin mathsize 14px style square root of V subscript x superscript 2 space plus space V subscript y superscript 2 end root space end style= 1.011 × 108 m/s
Angle made by the direction of Be nucleus with x-axis ,  tan-1 ( 6.455 / 7.787 ) =39.6o ( clockwise direction )
Answered by Thiyagarajan K | 22 Jul, 2022, 10:58: AM
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