By using the following atomic masses : `._(92)^(238)U = 238.05079u`. `._(2)^(4)He = 4.00260u, ._(90)^(234)Th = 234.04363u`.
`._(1)^(1)H = 1.007834, ._(91)^(237)Pa = 237.065121u` (i) Calculate the energy released during the `alpha-`decay of `._(92)^(238)U`.
(ii) Show that `._(92)^(238)U` cannot 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.