CBSE Class 11-science Answered

Dear student

Consider the neutron mass as   =  m

Corbon nucleus mass =  12m

If two masses m1, m2 collide head on with initial velocities  u1, u2 

the final velocities ofthe bodies are given by

In the problem 

Let ki  and kf are initial and final kinetc energies of   Neutron 

(a) 

% of Kinetic energy  transfer  to therest mass  =  (ki-kf)/ki  

= ( 4  m1  m2 ) / (m1+m2)² 

 =  (4X mX 12m )   /(  m+12m) ²   

=48m²/  169 m² =  0.284

One can derive the % of kinetic energy trans fer from the velociti equations given above.

The fraction of the neutron’s kinetic energy is transferred to the carbon nucleus is 0.284

(b)initial kinetic energy of Neutron = 1.60 X10 ¹³ J

FInal kinetic energy of Neutron =(1-0.284)  X 1.60 X10 ¹³   = 1.14 X10¹³  J

kinetic energy of the carbon nucleus after the collision = KE loss by Neutron

= 0.284  X 1.60 X10 ¹³   =0.454X10¹³  J

kinetic energy of the carbon nucleus after the collision = KE loss by Neutron =0.454X10¹³  J

Regards

Team

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