Question

Show that the energy released when three alpha particles fuse to form${}^{\mathbf{12}}\mathbf{C}$is$\mathbf{7}\mathbf{.}\mathbf{27}\mathbf{}\mathit{M}\mathit{e}\mathit{V}$. The atomic mass ofrole="math" localid="1661752248572" ${}^{\mathbf{4}}\mathit{H}\mathit{e}$is$\mathbf{4}\mathbf{.}\mathbf{0026}\mathit{u}$, and that of${}^{\mathbf{12}}\mathit{C}$is 12.0000u.

Short answer

The energy released for the fusion of three alpha particles to form ${}^{12}C$is 7.27 MeV.

Step-by-step solution

The given data

1. Atomic mass of ${}^{4}He$,$m_{4_{He}}=4.0026u$
2. Atomic mass of ${}^{12}C$,$m_{{12}_C}=12.0000u$

Determine the formula for energy:

The Q-value for a fusion process with energy being released is as follows:

$Q=∆m{c}^2$ …… (i)

Here,$c^2=931.5\frac{\mathrm{Me}V}{u}$

Determine the energy released for the fusion reaction

If$m_{4_{\mathrm{He}}}$is the mass of an atom of helium and$m_{{12}_C}$is the mass of an atom of carbon, then the energy released in a single fusion event is given using equation (i) as follows:

$Q=\left(3{m_4}_{He}-m_{{12}_C}\right)c^2$

Substitute the values and solve as:

$$\begin{gathered} Q=\left(3\left(4.0026u\right)-12.0000u\right)\left(931.5\frac{MeV}{N}\right) \\ =7.27MeV \end{gathered}$$

Hence, the energy released for the fusion to form ${}^{12}C$is 7.27 MeV.