JEE Mains · Physics · STD 12 - 13. Nuclei
The energy released per fission of nucleus of \({ }^{240} X\) is \(200 MeV\). The energy released if all the atoms in \(120 g\) of pure \({ }^{240} X\) undergo fission is \(........\times 10^{25}\,MeV\).\(\left(\right.\) Given \(\left.N_A=6 \times 10^{23}\right)\)
- A \(5\)
- B \(4\)
- C \(6\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(\text { No. of mole }=\frac{120}{240}=\frac{1}{2}\) \(\text { No. of molecules }=\frac{1}{2} \times N _{ A }\) \(\text { Energy released }=\frac{1}{2} \times 6 \times 10^{23} \times 200\) \(=6 \times 10^{25}\,MeV\)
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