If three helium nuclei combine to form a carb | vedclass

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Question

(C) The nuclear reaction is: $3 \, _2^4	ext{He} \longrightarrow _6^{12}	ext{C} + Q$.The mass of three helium nuclei is $3 	imes 4.002603 	ext{ u}

Vedclass · Accepted Answer

$727$

Vedclass · Answer

(C) The nuclear reaction is: $3 \, _2^4	ext{He} \longrightarrow _6^{12}	ext{C} + Q$.The mass of three helium nuclei is $3 	imes 4.002603 	ext{ u} = 12.007809 	ext{ u}$.The mass of one carbon nucleus is $12.000000 	ext{ u}$.The mass defect $\Delta m$ is calculated as: $\Delta m = (3 	imes m_{	ext{He}}) - m_{	ext{C}} = 12.007809 	ext{ u} - 12.000000 	ext{ u} = 0.007809 	ext{ u}$.The energy released $Q$ is given by: $Q = \Delta m 	imes 931 	ext{ MeV/u}$.$Q = 0.007809 	imes 931 	ext{ MeV} \approx 7.270179 	ext{ MeV}$.Expressing this in terms of $10^{-2} 	ext{ MeV}$,we get $727.0179 	imes 10^{-2} 	ext{ MeV} \approx 727 	imes 10^{-2} 	ext{ MeV}$.

If three helium nuclei combine to form a carbon nucleus,then the energy released in this reaction is $\times 10^{-2} \text{ MeV}$. (Given $1 \text{ u} = 931 \text{ MeV}/c^2$,atomic mass of helium $= 4.002603 \text{ u}$,atomic mass of carbon $= 12.000000 \text{ u}$)

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