Q.

A stationary Pb200 nucleus emits an alpha –particle with kinetic energy Tα. The fraction of recoil energy of the daughter nucleus to the total energy liberated is :

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a

1/196

b

4/196

c

1/20

d

1/50

answer is D.

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Detailed Solution

Pb200→X196+2H4 0=4V1+196V2 V1=1964V2 KT=12×4×V12+12×196V22 V1=−49V2 KD=12×196×V22 KDKT=196V224V12+196V22=49V22V12+49V22 49V22V22[(V12/V22)+49]=49(49)2+49 =149+1=150
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