You are given that mass of \({ }_a^7 \mathrm{Li}=7.0160 ~\text{u}\) Mass of \({ }_2^4 \mathrm{He}=4.0026 ~\text{u}\) and Mass of \({ }_1^1 \mathrm{H}=1.0079 ~\text{u}\)
When \(20~\text{g}\) of \({ }_a^7 \mathrm{Li}\) is converted into \({ }_2^4 \mathrm{He}\) By proton capture, the energy liberated, (in kWh), is: 
[Mass of nucleon = \(1~\text{GeV/c}^2\)]
1. \( 1.33 \times 10^6 \)
2. \( 8 \times 10^6 \)
3. \( 6.82 \times 10^5 \)
4. \( 4.5 \times 10^5 \)

Subtopic:  Nuclear Energy |
Level 3: 35%-60%
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The average energy released per fission for the nucleus of \({ }_{92}^{235} \mathrm{U}\) is \(190\) MeV. When all the atoms of \(47\) g pure \({ }_{92}^{235} \mathrm{U}\) undergo fission process, the energy released is \(\alpha \times 10^{23}\) MeV. The value of \(\alpha\) is:
(Avogadro Number \(= 6\times 10^{23}\) per mole)
1. \(220\)
2. \(228\)
3. \(210\)
4. \(230\)
Subtopic:  Nuclear Energy |
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