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JEE Mains · Physics · STD 12 - 13. Nuclei

You are given that Mass of \({ }_{3}^{7} Li =7.0160\, u\) Mass of \({ }_{2}^{4} He =4.0026\, u\) and Mass of \({ }_{1}^{1} H =1.0079\, u\) When \(20\, g\) of \({ }_{3}^{7} Li\) is converted into \({ }_{2}^{4} He\) by proton capture, the energy liberated, (in \(kWh\) ), is: [Mass of nudeon \(\left.=1\, GeV / c ^{2}\right]\)

  1. A \(8 \times 10^{6}\)
  2. B \(1.33 \times 10^{6}\)
  3. C \(6.82 \times 10^{5}\)
  4. D \(4.5 \times 10^{5}\)
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Answer & Solution

Correct Answer

(B) \(1.33 \times 10^{6}\)

Step-by-step Solution

Detailed explanation

\({ }_{3}^{7} Li +{ }_{1}^{1} H \rightarrow 2\left({ }_{2}^{4} He \right)\) \(\Delta m \Rightarrow\left[ m _{ Li }+ m _{ H }\right]-2\left[ M _{ He }\right]\) Energy released in \(1\) reaction \(\Rightarrow \Delta mc ^{2}\) In use of \(7.016\, u\) \(Li\) energy is…
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