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Questions  

A star initially has 3×1020  deuterons. It produces energy via the process   1H2+1H21H3+p and  1H2+1H32He4+n . Find the total energy released during the deuteron supply of star is exhausted (mass of  2He4=4.001amu,1H1=1.007amu,1H2=2.014amu,0n1=1.008amu )

a
1.1×108J
b
3.8×108J
c
2.8×108J
d
1.8×108J

detailed solution

Correct option is B

1H2+1H2→1H3+pplus 1H2+1H3→2He4+ngives31H2→2He4+n+pMass defect =ΔM=mass  of   2He4+n+P−mass 31H2 ⇒ΔM=4.001+1.008+1.007−32.014⇒ΔM=−0.026 amu∴Energy released =ΔE=ΔM931MeV=0.026×931×106×1.6×10−19J =3.87×10−12J No. of reactions =N3=3×10203=1020∴Total energy released =E=N3ΔE=3.87×10−12×1020=3.87×108J

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