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The nucleus of an atom is .92^235Y, init...

The nucleus of an atom is `._92^235Y`, initially at rest, decays by emitting an `alpha`-particle as per the equation
`._92^235Y to ._90^231X + ._2^4He`+ 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 of `alpha`-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 `alpha`-particle . Take mass of `alpha`-particle to be `6.68xx10^(-27)` kg.

Text Solution

Verified by Experts

Q `= [(7.835xx231 )+(7.07xx4)-(7.8xx235)]` MeV
=5.18 MeV
`=5.18 xx1.6 xx10^(-13)` J
This entire kinetic energy is taken by `alpha`-particle as given
`1/2 mv^2=5.18 xx1.6 xx10^(-13)`
` 1/2 xx6.68 xx10^(-27)v^2=5.18xx1.6 xx10^(-13)`
`v =1.57xx10^7` m/s
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