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

Hint: The Q-value of a reaction is given by \(Q=\Delta mc^2\).
Step 1: Find the fission reaction of \({ }_{26}^{56} \mathrm{Fe}\).
The fission of  F2656e can be given as:
\(_{26}^{56}\mathrm{Fe} \rightarrow 2{ }_{13}^{28} \mathrm{Al}\)

Step 2: Find the Q-value of the fission reaction.
It is given that:
Atomic mass of  mF2656e=55.9349 u
Atomic mass of  mA1328l=27.98191 u
The Q-value of this nuclear reaction is gien as:
Q=mF2656e-2mA1328lc2
=55.93494-2×27.9819c2
=--0.02886 c2u
But 1 u=931.5 MeV/c2
Q=-0.0288×931.5=-26.8272 MeV
The Q-value of the fission is negative. Therefore, fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.