We are given the following atomic masses:
`""_(92)^(238) U = 238.05079 u " " _(2)^(4)He = 4.00260 u `
`""_(90)^(234)Th = 234.04363 u" "_(1)^(1)H= 1.00783 u`
`""_(91)^(237)Pa = 237.05121 u`
Here the symbol Pa is for the element protactinium (Z = 91).
(a) Calculate the energy released during the alpha decay of `""_(92)^(238)U`.
(b) Show that `""_(92)^(238)U` can not spontaneously emit a proton.

Answer

(i) `Q = 4.25 MeV`
(ii) `Q = (-0.022165) ._(u)C^(2)`
(impossible)

Explanation

`._(92)^(238)U rarr_(90)^(234)Th+_(2)^(4)He`
`Deltam=(238.05079-4.00260-234.0 4363)u`
`E=Deltamc^(2)=4.24764 MeV`
If it emits proton spontaneously, the equation is not balanced in terms of atoms & mass number.
`._(92)^(238)U rarr_(91)^(237)Pa+_(1)^(4)H`
`Deltam'= (238.05079-237.065121-1.00784)u`
`=- 0.022165 u`
`:' Deltam` is negative, so reaction is not spontaneous.