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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`. |
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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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