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rajan jha Grade: 12
        

A nucleus X initially at rest , undergoes alpha decay acc. to the equation
232(X)Z--->A(Y)90 +ALPHA.[232 is mass no. ,Z is atomic no.same for Y].
What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle ?
{ans:-228/232}

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

let E amount of energy is released in this process then this energy is converted to kinetic energy of alfa particle & daughter nuclie...


initially the system is in rest so momentam = 0


 


finally momentam = PA + PD


PA = momentam of alfa particle & PD is momentam of daughter nuclie ....


here we can apply conservation of momentam so


PA + PD = 0 ................1


total kinetic energy = (KE)A + (KE)D = E .............2


kinetic energy(KE) = p2/2m so eq 1


(KE)AMA = (KE)DMD  .....................3


from eq 3 & 2


(KE)A + (KE)AMA/MD = E


 (KE)A(MA+MD)/MD = E


(KE)A/E = MD/(MA+MD)=228/232      


this is the required fraction


 

6 years ago
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