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The nucleus of an atom of .(92)Y^(235) i...

The nucleus of an atom of `._(92)Y^(235)` initially at rest decays by emitting an `alpha` particle. The binding energy per nucleon of parent and daughter nuclei are `7.8MeV` and `7.835MeV` respectively and that of `alpha` particles is `7.07MeV//"nucleon"`. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share of energy in the reaction, calculate speed of emitted alpha particle. Take mass of `alpha` particle to be `6.68xx10^(-27)kg`.

Text Solution

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The nuclear reaction is `._(92)Y^(235)to._(90)X^(231)+._2He^(4)+Q`
Total B.E. of `._(92)Y^(235)=7.89xx235Mev=1833MeV`
Total B.E. of `._(90)X^(231)=7.835xx231Mev=1809.9MeV`
B.E. of `._(2)He^(4)=7.07xx4=28.3MeV`
`:.` Q=Total BE of products-BE of parent atom=`1809.9+28.3-1833=5.2MeV`
`=5.2xx1.6xx10^(-13)J=8.32xx10^(-13)J`
This is the energy carried by `alpha` particle.
form `1/2mv^(2)=Q=8.32xx10^(-13)`
`v=sqrt((8.32xx2xx10^(-13))/m)=sqrt((16.64xx10^(-13))/(6.68xx10^(-27)))=1.58xx10^(7)m//s`
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