Atomic Physics Question 11
Question 11 - 2024 (31 Jan Shift 1)
The mass defect in a particular reaction is $0.4 \mathrm{~g}$. The amount of energy liberated is $\mathrm{n} \times 10^{7} \mathrm{kWh}$, where $\mathrm{n}=$[^0]
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Answer: (1)
Solution:
$$ \begin{aligned} & \mathrm{E}=\Delta \mathrm{mc}^{2} \ & =0.4 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2} \ & =3600 \times 10^{7} \mathrm{kWs} \ & =\frac{3600 \times 10^{7}}{3600} \mathrm{kWh}=1 \times 10^{7} \mathrm{kWh} \end{aligned} $$
