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 \(\left.=1\, GeV / c ^{2}\right]\)
- A \(8 \times 10^{6}\)
- B \(1.33 \times 10^{6}\)
- C \(6.82 \times 10^{5}\)
- D \(4.5 \times 10^{5}\)
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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