The nucleus of an atom `mu |_(92)^(235) Y |` , initially at rest, decays by emitting an a-particle as per the equation `""_(92)^(235)Yrarr""_(90)^(231)X+ ""_(2)^(4)` HC + Energy
It is given that the binding energies per nucleon of the parent and the daughter nuclei are 7.8 MeV and 7.835 MeV respectively and that ofc-particle is 7.07 MeV/nucleon. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, calculate the speed of the emitted co-particle. Take, mass of particle to be `6.68 xx 10^(-27)` kg.

Explanation

`""_(92)Y^(235) rarr ""_(90)X^(231) + ""_(2)He^(4)` + Energy
BE of Y = 235 `xx 7.8 = 1833`
BE of X = `231 xx 7.835 = 1809.88`
BE of He `= 4 xx 7.07 = 28 .28` MeV
Energy = BE (X +He) - BE (Y)
`=(1809.88+28.88) - 1833`
= 5.76 MeV
`= 5.76 xx 16 xx 10^(-13) `J
E = `922 xx 10^(-13)` J
`KE= (1)/(2) mv^(2)`
`922 xx 10^(-13) =(1)/(22) xx6.68 xx 10^(-27) xx v^(2)`
`v^(2) = (922 xx 10^(-13)xx2)/(6.68xx10^(-27))`
`v^(2) = 2.7 xx 10^(14) `
`v = 1.6 xx 10^(7) ` m/s