Question

Calculate the Q-value of the fusion reaction ${}^{4}He{+}^{4}He{=}^{8}Be$ Is such a fusion energetically favorable? Atomic mass of ${}^{8}Be$ is 8.0053 u and that of ${}^{4}He$ is 4.0026 u.

• A. The Q value is negative and the reaction is energetically favorable
• B. The Q value is positive and the reaction is energetically favorable
• C. The Q value is negative and the reaction is not energetically favorable
• D. The Q value is positive but the reaction is not energetically favorable

Answer 1

The correct option is C The Q value is negative and the reaction is not energetically favorable

First of all lets calculate the Q value of the reaction which is
Q-value =Rest mass energy of reactants -Rest mass energy of products.
$=[2\times m(He)-m(Be\left)\right]c^2.=\left[\right(2\times 4.0026)-8.0053]c^2=(8.0052u-8.0053u)c^2=(-0.0001u)c^2=(931\times -0.0001)Mev.=-0.0931Mev=-93.1keV$
We find that the Q value is negative which also implies that the rest mass energy in the reactant side is more than that on the product side.
So the energy is not being released but being consumed in the process and hence is energetically not favorable.