The mass defect in a particular nuclear reaction is $0.3 \mathrm{~g}$. The amount of energy liberated in…

The mass defect in a particular nuclear reaction is $0.3 \mathrm{~g}$. The amount of energy liberated in kilowatt hour is : (Velocity of light $=3 \times 10^8$ )
  1. $1.5 \times 10^6$
  2. $2.5 \times 10^6$
  3. $3 \times 10^6$
  4. $7.5 \times 10^6$

Solution

$\Delta m=0.3 \mathrm{~g}$ $=0.3 \times 10^{-3} \mathrm{~kg}=3 \times 10^{-4} \mathrm{~kg}$ Energy liberated, $E=\Delta m c^2$ $=3 \times 10^{-4} \times\left(3 \times 10^8\right)^2$ $=3 \times 10^{-4} \times 9 \times 10^{16}$ $=27 \times 10^{12} \mathrm{~J}$ $=\frac{27 \times 10^{12}}{3.6 \times 10^6} \mathrm{kWh}$ $=7.5 \times 10^6 \mathrm{kWh}$

Asked in: AP EAMCET 2003

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