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.

Explanation

(a) The alpha decay of `""_(92)^(238)U` is given by Eq.The energy released in this process is given by
`Q = (m_u - m_(He))c^2`
Substituting the atomic masses as given in the data. We find
`Q = (238.05079 - 234.04363 - 4.00260) u xx c^2`
`= (0.00456 u)c^2`
` = (0.00456 u)(931.5 MeV//u)`
`= 4.25 MeV`.
(b) If `""_(92)^(238)U` spontaneously emitts a proton, the decay process would be
`""_(92)^(238)U to ""_(91)^(237)U + ""_(1)^(1)H`
The Q for this process to happen is
`= (m_U - m_(Pa) - m_(H)) c^2`
`=(238.05079 - 237.05121 - 1.00783) u xx c^2`
`= (-0.00825 u)c^2`
`= -(0.00825 u)(931.5 MeV//u)`
`= -7.68 MeV`.
Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of `7.86 MeV` to a `""_(92)^(238)U` nucleus to make it emit a proton.