Question

For the carbon cycle to become established, helium-4 nuclei must fuse to form beryllium-8. Calculate Q for this reaction.

Short answer

The value of$Q=-0.08756MeV$

Step-by-step solution

Given data and concept

The nuclear reaction for this process:

${}_2{}^{4}He+{}_2{}^{4}He\to {}_4{}^{8}Be+\gamma$

Mass of helium =$4.002603u$

Mass of beryllium =$8.005305u$

The energy released in this decay depends upon the mass difference between the parent and daughter nucleus.

We can calculate the kinetic energy Q of the emitted particle using the formula:

$Q=\left(m_p-m_D\right)c^2$

In this formula${\text{m}}_P$ is the mass of the parent nucleus. And ${\text{m}}_D$is the mass of the daughter nucleus.

Finding Q

By substituting numerical values, we have:

$$\begin{gathered} Q=\left(2m_{He}-m_{Be}\right)c^2 \\ Q=\left(2\times 4.002603u-8.00530u\right)931.5\frac{MeV}{u} \\ Q=-0.08756MeV \end{gathered}$$

Therefore, the value of$Q=-0.08756MeV$