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Asked by Prashant DIGHE 16th January 2020, 10:01 PM
( It is given in the question that energy of β-particle emitted is 6.7 MeV.
This is assumed as typo error. It is considered that energy of α-particle emitted is 6.7 MeV )

Energy of daughter nucleus is obtained from conservation of momentum.

If we assume the parent nuclei is initially at rest, then we have,  mα × vα + M × V = 0  ......................(1)

where mα is mass of α-particle , vα is the speed of α-particle, M is mass of daughter nucleus and V is speed of daughter nucleus.

If E is the energy of α-particle, then its speed is obtained from,  (1/2)mα vα2 = E  ,   then vα = (2E/mα )1/2

Hence Speed of daughter nucleus , V = - ( mα × vα ) / M   = - [ mα × (2E/mα )1/2 ] / M

Hence recoil energy of daughter nucleus = (1/2) M V2 = ( mα × E ) / M  = ( 4 × 6.7 )/214 = 0.125 MeV
Answered by Expert 17th January 2020, 8:57 AM
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