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If in a nuclear fission, piece of uranium of mass 0.5 g is lost, the energy obtained in kWh is

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$1.25 x 10^{7}$

$2.25 x 10^{7}$

$3.25 x 10^{7}$

$0.25 x 10^{7}$

answer is A.

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Detailed Solution

$E = ∆ {mc}^{2} = 0.5 x 10^{- 3} x {(3 x 10^{8})}^{2} = 4.5 x 10^{13} J \Rightarrow E = \frac{4.5 x 10^{13}}{3.6 x 10^{6}} = 1.25 x 10^{7} kWh .$

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