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Standard XII

Physics

Nuclear Force

The binding e...

Question

The binding energy per nucleon for $U^{238}$ about 7.5Mev where as it is about 8.5 Mev for a nucleus having a mass half of Uranium. If $U^{238}$ splits into two exact halves the energy released would be

6.4 Mev

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119 Mev

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238 Mev

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476 Mev

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Solution

The correct option is A 238 Mev
$\begin{matrix} Here, energy released is due to mass defect. U^{238} ⟶ 2 (X^{119}) \end{matrix}$
$= 238(8.5)-2(119)(7.5) MeV$
$Energy released is 238 MeV.$

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The binding energy per nucleon for U238 about 7.5Mev where as it is about 8.5 Mev for a nucleus having a mass half of Uranium. If U238 splits into two exact halves the energy released would be

The correct option is A 238 Mev Here, energy released is due to mass defect. U238⟶2(X119)= 238(8.5)-2(119)(7.5) MeV Energy released is 238 MeV.

The binding energy per nucleon for $U^{238}$ about 7.5Mev where as it is about 8.5 Mev for a nucleus having a mass half of Uranium. If $U^{238}$ splits into two exact halves the energy released would be

The correct option is A 238 Mev
$\begin{matrix} Here, energy released is due to mass defect. U^{238} ⟶ 2 (X^{119}) \end{matrix}$
$= 238(8.5)-2(119)(7.5) MeV$
$Energy released is 238 MeV.$