Q168 / 170 · Manipal (MET) 2018 · Physics

The mass defect in a particulạr nuclear reaction is 0.3 g . The amount of energy liberated in killowatt hour is (velocity of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ )

  • A
    $1.5 \times 10^6$
  • B
    $2.5 \times 10^6$
  • C
    $3 \times 10^6$
  • Correct answer
    D
    $7.5 \times 10^6$

Solution

$\begin{aligned} E & =\Delta m c^2 \\ E & =\frac{0.3}{1000} \times\left(3 \times 10^8\right)^2 \\ & =2.7 \times 10^{13} \mathrm{~J} \\ & =\frac{2.7 \times 10^{13}}{3.6 \times 10^6}=7.5 \times 10^6 \mathrm{kWh}\end{aligned}$

Hence, the correct answer is option: 'D'.

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