Suppose we think of the fission of a $^{56}_{26} Fe$ nucleus into two equal fragments,$^{28}_{13} Al$. Is the fission energetically possible? Argue by working out the $Q$-value of the process.
Given: $m(^{56}_{26} Fe) = 55.93494 \; u$ and $m(^{28}_{13} Al) = 27.98191 \; u$.

(N/A) The fission of $^{56}_{26} Fe$ can be represented as:
$^{56}_{26} Fe \rightarrow 2 \; ^{28}_{13} Al$
Given:
Atomic mass of $m(^{56}_{26} Fe) = 55.93494 \; u$
Atomic mass of $m(^{28}_{13} Al) = 27.98191 \; u$
The $Q$-value of this nuclear reaction is calculated as:
$Q = [m(^{56}_{26} Fe) - 2 \times m(^{28}_{13} Al)] \times c^2$
$Q = [55.93494 - 2 \times 27.98191] \; u \times c^2$
$Q = [55.93494 - 55.96382] \; u \times c^2$
$Q = -0.02888 \; u \times c^2$
Since $1 \; u = 931.5 \; MeV/c^2$:
$Q = -0.02888 \times 931.5 \; MeV$
$Q = -26.902 \; MeV$
Since the $Q$-value is negative,the fission is not energetically possible. For a fission reaction to be spontaneous or energetically possible,the $Q$-value must be positive.