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A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of " "_(63)^(29)Bi atoms (of mass 62.92960 u). |
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Answer» Solution :It is GIVEN that `m_(H)` = 1.007825 u, `m_(n)` = 1.008665 u and `m_(Cu)` = 62.92960 u Every atom of `" "_(29)^(63)Cu` contains 29 protons, 29 electrons (i.e., EFFECTIVELY 29 H atom) and (63 - 29) = 34 neutrons. We know that 1 mole of copper WEIGHS 63 g and contains `N_(A) = 6 xx 10^(23)` atoms of copper (or `29 xx 6XX 10^(23) m_(H)` and `34 xx 6 xx 10^(23)`neutrons). Hence, corresponding number in 3 g of copper will be : `n_(p) = 3/63xx29xx6xx10^(23), n_(n)= 34xx 6xx 10^(23)` and `n_(Cu) = 3/63xx6xx 10^(23)` `THEREFORE` Total mass defect `DeltaM = n_(p). m_(H) + n_(n). m_(n) + n_(Cu).m_(Cu)= (3xx29xx6xx10^(23))/63xx1.007825+(3xx34xx6xx10^(23))/63xx1.008665-(3xx6xx10^(23))/63xx 62.92960 u=(6xx10^(23))/21[29xx1.007825+34xx1.008665-62.92960]u=(6xx10^(23))/21xx0.591935u` `therefore` Total energy required to split 3.0g of Cu into protons and neutrons `E=DeltaMxx931.5MeV=(6xx10^(23)xx0.591935)/21xx931.5MeV=1.584xx10^(25)MeV=2.535xx10^(12)`J. |
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