Calculate the energy released in $m e V$ in the following nuclear reaction:
$_{92}^{298} U ⟶_{90}^{234} T H +_{2}^{4} H e + Q$
[Mass of $_{92}^{298} U = 238.05079 u$ Mass of $_{90}^{234} T H = 234.043630 u]$

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Solution

Given:
Mass of $_{92}^{298} U = 238.05079 u$ , Mass of $_{90}^{234} T H = 234.043630 u$

Mass Defect = Mass of uranium - Mass of thorium - Mass of helium

$Mass defect = 238.05079 u - 234.043630 u - 4.002600 u = 0.004564 u$

Now,
$1 u releases 931.5 M e V / c^{2} energy$

$\Rightarrow$ Energy released by $0.004564 u$ = $(0.00456 \times 931.5) M e V / c^{2} = 4.24764 M e V / c^{2}$

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