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We are given the following atomic masses...

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

Text Solution

Verified by Experts

If `._(92)^(238)U` spontaneously emits a proton, the decay process would be
`._(92)^(238)U to ._(91)^(237)Pa + ._(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 7.68 MeV to a `._(92)^(238)U` nucleus to make it emit a proton.
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