Question

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`.

Answer 1

Q=[(7.835 x 231) + (7.07 x 4) - (7.8 x 235) ] MeV
=5.18 MeV
=`5.18xx1.6xx10^(-13)` J
This entire kinetic energy is taken by `alpha`-particle as given
`=1/2mv^2=5.18xx1.6xx10^(-13)`
`=1/2xx6.68xx10^(-27)v^2=5.18xx1.6xx10^(-13)`
`v=1.57xx10^7` m/s