Question 28.QE.6: Energy from a fission reaction Determine the energy released......
Energy from a fission reaction
Determine the energy released in the above fission reaction. Use the following masses: { }_0^1 \mathrm{n}(1.008665 \mathrm{u}),{ }_{92}^{235} \mathrm{U}(235.043924 \mathrm{u}),{ }_{56}^{141} \mathrm{Ba}(140.914411 \mathrm{u}),{ }_{362}^{92} \mathrm{Kr}(91.926156 \mathrm{u}). . One neutron collides with and joins momentarily with the U-235 atomic nucleus to form an excited U-236, which quickly splits into Ba-141, Kr-92, and three neutrons.
Represent mathematically The energy Q released in the reaction is the energy equivalent to the mass difference between the reactants and products.
Q=\left[m\left({ }_0^1 \mathrm{n}\right)+m\left({ }_{92}^{235} \mathrm{U}\right)\right] c^2
-\left[m\left({ }_{56}^{141} \mathrm{Ba}\right)+m\left({ }_{36}^{92} \mathrm{Kr}\right)+3 m\left({ }_0^1 \mathrm{n}\right)\right] c^2
=\left[m\left({ }_{92}^{235} \mathrm{U}\right)-m\left({ }_{56}^{141} \mathrm{Ba}\right)-m\left({ }_{36}^{92} \mathrm{Kr}\right)-2 m\left({ }_0^1 \mathrm{n}\right)\right] c^2
Solve and evaluate Inserting the appropriate values and converting to MeV, we get
Q=[235.043924 \mathrm{u}-140.914411 \mathrm{u}-91.926156 \mathrm{u} -2(1.008665 \mathrm{u})] c^2\left(\frac{931.5 \mathrm{MeV}}{\mathrm{u} \cdot c^2}\right)
= 173 MeV
This is a typical energy release during fission reactions. This reaction produces over 300 million times more energy than burning one octane molecule.
Try it yourself: Your friend says that nuclear power plants could release energy by the fission of iron-56 (very abundant) instead of the much less abundant uranium- 235. What do you say?
Answer: Iron-56 has about the highest binding energy per nucleon of any nucleus. This means energy must be added to the nucleus to cause it to break into smaller nuclei with lower BE/A—exactly the opposite of what is desired in a nuclear power plant.