The mass defect in a particular reaction is $0.4 \mathrm{~g}$. The amount of energy liberated is $n \times 10^7 \mathrm{~kWh}$, where $n=$ __________. (speed of light $\left.=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$
Answer (integer)
1
Solution
<p>$$\begin{aligned}
& \mathrm{E}=\Delta \mathrm{mc}^2 \\
& =0.4 \times 10^{-3} \times\left(3 \times 10^8\right)^2 \\
& =3600 \times 10^7 \mathrm{kWs} \\
& =\frac{3600 \times 10^7}{3600} \mathrm{kWh}=1 \times 10^7 \mathrm{kWh}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Mass-Energy Equivalence
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