In the reaction $4 \mathrm{He}+{ }^{9} \mathrm{Be} \rightarrow \mathrm{n}+{ }^{12} \mathrm{C}$, how much energy is released or absorbed? The masses of the neutral atoms are as follows: the mas of $4 \mathrm{He}$ is $4.0026033 \mathrm{u}$, the mass of ${ }^{9} \mathrm{Be}$ is $9.0121831 \mathrm{u}$, the mass of $\mathrm{n}$ is $1.0086649 \mathrm{u}$, and the mass of $12 \mathrm{C}$ is $12.0000000 \mathrm{u}$.

Question

Quizplus · Accepted Answer

To calculate the energy released or absorbed, we first find the mass difference before and after the reaction. The total mass before the reaction is $4.0026033 \mathrm{u} + 9.0121831 \mathrm{u} = 13.0147864 \mathrm{u}$. The total mass after the reaction is $1.0086649 \mathrm{u} + 12.0000000 \mathrm{u} = 13.0086649 \mathrm{u}$. The mass difference is $13.0147864 \mathrm{u} - 13.0086649 \mathrm{u} = 0.0061215 \mathrm{u}$. Converting this mass difference to energy using Einstein's equation $E=mc^2$ and considering $1 \mathrm{u} = 931.5 \mathrm{MeV/c^2}$, the energy released is $0.0061215 \mathrm{u} \times 931.5 \mathrm{MeV/u} = 5.70 \mathrm{MeV}$.

In the Reaction 4He+9Be→n+12C4 \mathrm{He}+{ }^{9} \mathrm{Be} \rightarrow \mathrm{n}+{ }^{12} \mathrm{C}4He+9Be→n+12C

In the Reaction $4 \mathrm{He}+{ }^{9} \mathrm{Be} \rightarrow \mathrm{n}+{ }^{12} \mathrm{C}$