1.

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 dougther nuclei are 7.8MeV and 7.835MeV respectively and that of `alpha` particles is `7.07MeV//"nucleon"`. Assuming the dougther 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» Correct Answer - `1.573xx10^(7)m//s`
The given reaction is
`._(92)Y^(235) to ._(90)X^(231)+._(2)He^(4) +"energy"`
Energy released during the decay process is
`E=7.835xx231+4xx7.07-7.8xx235`
`E=1809.885+28.28-1833.3=5.165MeV`
`5.165xx1.6xx10^(-13)J`
If v is speed of alpha particle emitted, then
`1/2mv^(2)=E`
`v=sqrt((2E)/m)`
`=sqrt((2xx5.165xx1.6xx10^(-13))/(6.68xx10^(-27)))`
`=sqrt(16.528/6.68)xx10^(7)m//s`
`v=1.573xx10^(7)m//s`


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