A certain mass of Hydrogen is changed to Heli | vedclass

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(B) The fusion reaction is $4_{1}^{1} H ightarrow _{2}^{4} He + 2e^{+} + 2
u + Q$. The total mass defect for the formation of one Helium nucleus $(

Vedclass · Accepted Answer

$6.675$

Vedclass · Answer

(B) The fusion reaction is $4_{1}^{1} H ightarrow _{2}^{4} He + 2e^{+} + 2
u + Q$. The total mass defect for the formation of one Helium nucleus $(_{2}^{4} He)$ from four Hydrogen nuclei is $\Delta M = 0.02866 \, u$. The total energy released in this fusion process is $E = \Delta M 	imes 931 \, MeV$. $E = 0.02866 	imes 931 \approx 26.7 \, MeV$. Since the Helium nucleus has a mass number of $4$,the energy liberated per nucleon (or per $u$ of the mass involved in the product) is calculated by dividing the total energy by the mass number of the Helium nucleus. Energy per $u = \frac{26.7 \, MeV}{4} = 6.675 \, MeV$.

$A$ certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in the fusion reaction is $0.02866 \, u$. The energy liberated per $u$ is ........... $MeV$. (Given $1 \, u = 931 \, MeV$)

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