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`        A nucleus X initially at rest , undergoes alpha decay acc. to the equation232(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

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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