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Show that ""_(92)^(238)U cannot spontaneosly emit a proton. Given : ""_(92)^(238)U=238.05079u, ""_(91)^(237)Pa=237.05121u, ""_(1)^(1)H=1.00783u |
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Answer» Solution :The possible reaction is : `""_(92)^(238)U rarr ""_(91)^(237)Pa+""_(1)^(1)H` If the reaction takes place then Q - VALUE of the reaction is : Q- value `=[M_(U)-M)(Pa)-M_(H)]xxc^(2)` `[238.05079-237.05121-1.00783]uxxc^(2)` `=[-0.00825u]xx931.5(MeV)/(u)=-7.68MeV` As Q - value is `-ve`, the reaction is ENDOTHERMIC and NEEDS energy to be supplied. So, the reaction cannot take place spontaneously by itself. |
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