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JEE Mains · Physics · STD 12 - 13. Nuclei

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\(\mathrm{n}=\)________. (speed of light \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\) )

  1. A \(10\)
  2. B \(1\)
  3. C \(5\)
  4. D \(11\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

\(\mathrm{E}=\Delta \mathrm{mc}^2\) \(=0.4 \times 10^{-3} \times\left(3 \times 10^8\right)^2\) \(=3600 \times 10^7 \mathrm{kWs}\) \(=\frac{3600 \times 10^7}{3600} \mathrm{kWh}=1 \times 10^7 \mathrm{kWh}\)
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