In a certain star- three alpha particles undergo fusion in a single reaction to form 126C nucleus- Calculate the energy released in this reaction in MeV-Given- m42He - 4-002604 u and m126C - 12-000000 u

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Standard XII

Physics

Binding Energy

In a certain ...

Question

In a certain star, three alpha particles undergo fusion in a single reaction to form $_{6}^{12} C$ nucleus. Calculate the energy released in this reaction in $M e V$ .
Given: $m (_{2}^{4} H e) = 4.002604 u$ and $m (_{6}^{12} C) = 12.000000 u$

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Solution

Mass of $3$ alpha particles $= 3 \times 4.002604 u$
$= 12.007812 u$ .
Mass of 1 $_{6} C^{12} a t o m = 12.007812 u$
Mass Defect $= △ m = 12.007812 - 12.000000 = 0.007812$
Energy released $= 0.007812 \times 931.5 = 7.2768 M e V$ .

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In a certain star, three alpha particles undergo fusion in a single reaction to form 126C nucleus. Calculate the energy released in this reaction in MeV.Given: m(42He)=4.002604 u and m(126C)=12.000000 u

Mass of 3 alpha particles =3×4.002604 u=12.007812 u.Mass of 1 6C12atom=12.007812 uMass Defect =△m=12.007812−12.000000=0.007812Energy released =0.007812×931.5=7.2768 MeV.

In a certain star, three alpha particles undergo fusion in a single reaction to form $_{6}^{12} C$ nucleus. Calculate the energy released in this reaction in $M e V$ .
Given: $m (_{2}^{4} H e) = 4.002604 u$ and $m (_{6}^{12} C) = 12.000000 u$

Mass of $3$ alpha particles $= 3 \times 4.002604 u$
$= 12.007812 u$ .
Mass of 1 $_{6} C^{12} a t o m = 12.007812 u$
Mass Defect $= △ m = 12.007812 - 12.000000 = 0.007812$
Energy released $= 0.007812 \times 931.5 = 7.2768 M e V$ .