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.^(238)U=238.05079 u" ".(2)^(4)He=4.002...

`.^(238)U=238.05079 u" "._(2)^(4)He=4.00260 u`
`._(90)^(234)Th=234.04363 n" "._(1)^(1)H=1.00783 u`
`._(90)^(237)Pa=237.0512l u`
Here the symbol `Pa` is for the element protacttntum `(Z=91)`.
(b) Show that `.^(238)U`. can not spontaneously emit a proton .

A

The energy released during the `alpha-` decay of `._(92)^(238)U` is `4.75MeV` approximately

B

The energy released during the `alpha-` decay of `._(92)^(238)U` is `4.25MeV` approximately

C

The emission of proton from `._(92)^(238)U` can be spontaneous

D

The emission of proton from `._(92)^(238)U` can not be spontaneous.

Text Solution

Verified by Experts

The correct Answer is:
B, D
The energy released in `alpha`- decay is given by
`Q=(M_(U)-M_(Th)-M_(He))c^(2)`
Substituting the atomic masses as given in the data, we find
`Q=(238.05079-234.04363-4.00260)uxxc^(2)=(0.00456u)c^(2)=(0.00456u)c^(2)`
`=(0.00456u)(931.5MeV//u)=4.25MeV`
If `._(92)^(238)U` spontaneously emits a proton, the decay process would be `._(92)^(238)Uto_(92)^(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)uxxc^(2)=(-0.00825u)c^(2)`
`=-(0.00825u)(931.5MeV//u)=-7.68MeV`
Thus, the `Q` of the process is negative and therefore it cannot proceed spiontaneously. We will have to supply an energy of `7.68MeV` to `._(92)^(238)U` nucleus to make it emit a proton.
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