The fission of Fe-56 into two fragments of $\,{}^{28}_{13}\mathrm{Al}$ with energy released Q can be written as $\,{}^{56}_{26}\mathrm{Fe}$ → $\,{}^{28}_{13}\mathrm{Al}$ + $\,{}^{28}_{13}\mathrm{Al}$ + Q Q = $\left[ m\left(\,{}^{56}_{26}\mathrm{Fe}\right)-2m\left(\,{}^{28}_{13}\mathrm{Al}\right)\right]c^2]$ = [55.93494 – 2 × 27.98191] × 931.5 MeV = – 0.02888 × 931.5 = – 26.90 MeV As the Q-value is negative, the fission is not possible energetically.