In a nuclear fusion process,the mass defect i | vedclass

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(C) The mass defect $\Delta m$ is given as $0.05\%$ of the total mass $m = 1 \, kg$.$\Delta m = \frac{0.05}{100} 	imes 1 \, kg = 5 	imes 10^{-4} \,

Vedclass · Accepted Answer

$45 	imes 10^{12} \, J$

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(C) The mass defect $\Delta m$ is given as $0.05\%$ of the total mass $m = 1 \, kg$.$\Delta m = \frac{0.05}{100} 	imes 1 \, kg = 5 	imes 10^{-4} \, kg$.The energy released $E$ is calculated using Einstein's mass-energy equivalence formula $E = (\Delta m) c^2$,where $c = 3 	imes 10^8 \, m/s$.$E = (5 	imes 10^{-4} \, kg) 	imes (3 	imes 10^8 \, m/s)^2$.$E = (5 	imes 10^{-4}) 	imes (9 	imes 10^{16}) \, J$.$E = 45 	imes 10^{12} \, J$.

In a nuclear fusion process,the mass defect is $0.05\%$. How much energy will be released from $1 \, kg$ of mass?

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