Question

Could a $56\mathrm{Fe}$nucleus fission into two $28\mathrm{Al}$nuclei? Your answer, which should include some calculations, should be based on the curve of binding energy.

Short answer

$∆E_2=470.4MeV$

negative sign of energy indicates that the energy is absorbed in the reaction. But in the fission reactions, energy releases. Hence, with the release of energy, ${}^{56}\mathrm{Fe}$cannot undergo fission into two ${}^{28}\mathrm{AI}$nuclei.

Step-by-step solution

Given Information

Here, $∆$$E$is energy released in the fission reaction.

From the figure localid="1651158118848" $42.6$, plot binding energy per nucleon verse mass number given in the text book, the binding energy per nucleon for localid="1651158127119" ${}^{56}\mathrm{Fe}$nucleus approximately is localid="1651158136858" $8.8\mathrm{MeV}$and the binding energy per nucleon for localid="1651158150390" $28\mathrm{Al}$nucleus is localid="1651158159913" $8.4\mathrm{MeV}$.

The nucleus localid="1651158165271" ${}^{56}\mathrm{Fe}$has localid="1651158172295" $56$nucleons, so the total binding energy of one nucleus is,

localid="1651158183077" $$\begin{gathered} ∆{\mathrm{E}}_1=\left(56\right)\left(8.8\mathrm{MeV}\right) \\ =492.8\mathrm{MeV} \\ \mathrm{The}\mathrm{nuclie}{}^{28}\mathrm{AI}\mathrm{nucleus}\mathrm{is}8.4\mathrm{MeV} \end{gathered}$$

Total binding energy two localid="1651158196999" ${}^{28}\mathrm{AI}$nucleus is,

localid="1651158204937" $∆E_2=470.4MeV$

Calculation

The amount of energy released in the nuclear reaction which involves a ⁵⁶ nucleus fission into two${}^{28}\mathrm{AI}$nuclei is

$$\begin{gathered} ∆E=\left(Totalbindingenergyoftw{o}^{28}AInuclei\right)- \\ \left(Totalbindingenergyof{}^{56}Fenucleus\right) \\ =470.4MeV-492.8MeV \\ =-22.4MeV \end{gathered}$$

Here, negative sign of energy indicates that the energy is absorbed in the reaction. But in the fission reactions, energy releases. Hence, with the release of energy, ${}^{56}\mathrm{Fe}$cannot undergo fission into two ${}^{28}\mathrm{AI}$nuclei.