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Find the energy Q liberated in beta^(-_(-)) and beta^(+)- decays and in K- caputure if the masses of the parent atom m_(p), the daughter atom M_(d) and an electron m are known. |
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Answer» Solution :In `beta^(-)` decay `Z^(X^(A))rarr_(Z+1)Y^(A)+e^(_)+Q` `Q=(M_(x)-M_(y)-m_(e ))C^(2)` `=[(M_(x)+ZM_(e )c^(2))-(M_(y)+Zm_(e )+m_(e ))]c^(2)` `=(M_(p)-M_(d))c^(2)` SINCE `M_(p),M_(d)` are the masses of the atoms. The binding energy of the ELECTRONS in ignored. In `K` capture `e_(k)^(-)+._(Z)X^(A) rarr_(z-1)Y^(A)+Q` `Q=(M_(X)-M_(Y))c^(2)+m_(e )c^(2)` `=(M_(x)^(c^(2))+Zm_(e )c^(2))-(M_(Y)c^(2)+(Z-1)m_(e )c^(2))` `=c^(2)(M_(p)-M_(d))` In `beta^(+) decay _(Z)X^(A) rarr_(Z-1)Y^(A)+e^(+)+Q` Then `Q= (M_(x)-M_(y)-m_(e ))c^(2)` `[M_(x)+Zm_(e )]c^(2)-[M_(y)+(Z-1)m_(e )]c^(2)-2m_(e )c^(2)` `=(M_(p)-M_(d)-2m_(e ))c^(2)` |
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