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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`-particle `._(92)^(238)U = 3.85395 xx 10^(-25) kg " " ._(90)^(234)Th = 3.78737 xx 10^(-25) kg " " ._(2)^(4)He = 6.64807 xx 10^(-27)` |
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Answer» Correct Answer - `8.8 xx 10^(-13)J` |
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