Determine the Q -value for the fusion reaction ^ 4 He + ^ 4 He → ^...

The $Q$-value of a nuclear reaction is given by the mass difference between reactants and products multiplied by $c^2$.

$Q = (\Delta m)c^2 = (2m_{\text{He}} - m_{\text{Be}})c^2$

Given $m_{\text{He}} = 4.0026 \text{ u}$ and $m_{\text{Be}} = 8.0053 \text{ u}$.

$\Delta m = 2(4.0026) - 8.0053$

$\Delta m = 8.0052 - 8.0053 = -0.0001 \text{ u}$

Converting this mass difference into energy using $1 \text{ u} \approx 931 \text{ MeV}$:

$Q = -0.0001 \times 931 \text{ MeV} = -0.0931 \text{ MeV} = -93.1 \text{ keV}$

Since the $Q$-value is negative, energy is absorbed in the reaction, making it an endothermic process. Thus, it is not energetically favourable.