What is the energy released in the process 3 | vedclass

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(A) The nuclear reaction is $3 \, _{2}^{4}	ext{He} 	o _{6}^{12}	ext{C}$.The mass defect $\Delta m$ is calculated as: $\Delta m = (3 	imes 	ext{ma

Vedclass · Accepted Answer

$7.27$

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(A) The nuclear reaction is $3 \, _{2}^{4}	ext{He} 	o _{6}^{12}	ext{C}$.The mass defect $\Delta m$ is calculated as: $\Delta m = (3 	imes 	ext{mass of } _{2}^{4}	ext{He}) - 	ext{mass of } _{6}^{12}	ext{C}$.$\Delta m = (3 	imes 4.002604 \, 	ext{amu}) - 12.000000 \, 	ext{amu}$.$\Delta m = 12.007812 \, 	ext{amu} - 12.000000 \, 	ext{amu} = 0.007812 \, 	ext{amu}$.Since $1 \, 	ext{amu} = 931.5 \, 	ext{MeV}/c^2$,the energy released $Q$ is:$Q = 0.007812 	imes 931.5 \, 	ext{MeV} \approx 7.27 \, 	ext{MeV}$.

What is the energy released in the process $3 \, _{2}^{4}\text{He} \to _{6}^{12}\text{C}$ (in $, \text{MeV}$)? (Mass of $_{2}^{4}\text{He} = 4.002604 \, \text{amu}$,mass of $_{6}^{12}\text{C} = 12 \, \text{amu}$)

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