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}

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 = P_A + P_D

P_A = momentam of alfa particle & P_D is momentam of daughter nuclie ....

here we can apply conservation of momentam so

P_A + P_D = 0 ................1

total kinetic energy = (KE)_A + (KE)_D = E .............2

kinetic energy(KE) = p²/2m so eq 1

(KE)_AM_A = (KE)_DM_{D  .....................3}

from eq 3 & 2

(KE)_A + (KE)_AM_A/M_D = E

 (KE)_A(M_A+M_D)/M_D = E

(KE)_A/E = M_D/(M_A+M_D)=228/232      

this is the required fraction