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(A) In the process of nuclear fusion, four hydrogen nuclei (protons) fuse to form one helium nucleus $(^4_2He)$.The reaction is given by: $4^1_1H \rig

Vedclass · Accepted Answer

$26.7$

Vedclass · Answer

(A) In the process of nuclear fusion, four hydrogen nuclei (protons) fuse to form one helium nucleus $(^4_2He)$.The reaction is given by: $4^1_1H ightarrow ^4_2He + 2e^+ + 2
u_e + 	ext{Energy}$.The mass defect $(\Delta m)$ for this reaction is approximately $0.0286 \ u$.Using the energy equivalence $E = \Delta m 	imes 931.5 \ MeV/u$, we get:$E = 0.0286 	imes 931.5 \approx 26.7 \ MeV$.Therefore, the energy released is $26.7 \ MeV$.

How much energy is released by the fusion of four hydrogen atoms to form a helium nucleus (in $MeV$)?

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