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(B) The energy released corresponding to the mass defect $\Delta m$ is given by $E = \Delta m 	imes 931 \, MeV$.Given $\Delta m = 0.02866 \, u$,the t

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$6.675$

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(B) The energy released corresponding to the mass defect $\Delta m$ is given by $E = \Delta m 	imes 931 \, MeV$.Given $\Delta m = 0.02866 \, u$,the total energy released for the formation of one helium nucleus (which has a mass of approximately $4 \, u$) is $E = 0.02866 	imes 931 \, MeV = 26.68246 \, MeV$.To find the energy released per $1 \, u$ of mass formed,we divide the total energy by the mass of the helium nucleus $(4 \, u)$:Energy per unit mass $= \frac{26.68246 \, MeV}{4 \, u} = 6.6706 \, MeV \approx 6.675 \, MeV$.

In a nuclear fusion process,a certain mass of hydrogen is converted into helium,resulting in a mass defect of $0.02866 \, u$. The energy released per $1 \, u$ of mass formed is ......... $MeV$. (Given $1 \, u = 931 \, MeV$)

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