Question 25.12: Helium Burning Find the energy released in the reaction 3^4H...
Helium Burning
Find the energy released in the reaction 3^{4} He \rightarrow{ }^{12} C .
ORGANIZE AND PLAN As usual with nuclear reactions, the mass difference between initial and final particles is large enough that we can use E=\Delta m c^{2} to find the energy release.
\text { Known: } M\left({ }^{4} He \right)=4.0026 u ; M\left({ }^{12} C \right)=12.0000 u ; 1 u \cdot c^{2}= 931.5 MeV.
For this process
E=\Delta m c^{2}=\left[3 M\left({ }^{4} He \right)\right] c^{2}-\left[M\left({ }^{12} C \right)\right] c^{2}.
=[3(4.0026 u )-12.0000 u ] c^{2}=0.0078 u \cdot c^{2}.
so
E=0.0078 u \cdot c^{2} \times \frac{931.5 MeV }{ u \cdot c^{2}}=7.27 MeV.
REFLECT Each helium-burning reaction produces less energy than a proton-proton reaction. However, helium burning occurs at a faster rate. When the Sun begins its helium-burning phase, the increased energy production will expand our star into a red giant.