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JEE MAIN 2026
23-01-26 S2
Question
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 $\alpha \times 10^{23} \mathrm{MeV}$. The value of $\alpha$ is $\_\_\_\_$。 (Avogadro Number $=6 \times 10^{23}$ per mole)
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JEE Main
Physics
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