The nucleus `._(92)^(235)U` decays according to `._(92)^(238)U rarr ._(92)^(234)Th + ._(2)^(4)He`. Calculate the kinetic energy of the emitted `alpha`

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Answer :

The nucleus `._(92)^(235)U` decays according to `._(92)^(238)U rarr ._(92)^(234)Th + ._(2)^(4)He`. ... kg " " ._(2)^(4)He = 6.64807 xx 10^(-27)`

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