What statement can be made about the masses of atoms in the above reaction?
A. \(m\left({ }_{92}^{235} \mathrm{U}\right)>m\left({ }_{56}^{141} \mathrm{Ba}\right)+m\left({ }_{36}^{92} \mathrm{Kr}\right)+2 m\left({ }_{0}^{1} \mathrm{n}\right)\)
B. \(m\left({ }_{92}^{235} \mathrm{U}\right)C. \(m\left({ }_{92}^{235} \mathrm{U}\right)=m\left({ }_{56}^{141} \mathrm{Ba}\right)+m\left({ }_{36}^{92} \mathrm{Kr}\right)+2 m\left({ }_{0}^{1} \mathrm{n}\right)\)
D. \(m\left({ }_{92}^{235} \mathrm{U}\right)=m\left({ }_{56}^{141} \mathrm{Ba}\right)+m\left({ }_{36}^{92} \mathrm{Kr}\right)+3 m\left({ }_{0}^{1} \mathrm{n}\right)\)

The uranium isotope \({ }^{235} \mathrm{U}\) can fission-break into two smaller-mass components and free neutrons-if it is struck by a free neutron. A typical reaction is

\[{ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n}\]

As you can see, the subscripts (the number of protons) and the superscripts (the number of nucleons) "balance" before and after the fission event; there is no change in the number of protons or neutrons. Significant energy is released in this reaction. If a fission event happens in a large chunk of \({ }^{235} \mathrm{U}\), the neutrons released may induce the fission of other \({ }^{235} \mathrm{U}\) atoms, resulting in a chain reaction. This is how a nuclear reactor works.

The number of neutrons required to create a stable nucleus increases with atomic number. When the heavy \({ }^{235} \mathrm{U}\) nucleus fissions, the lighter reaction products are thus neutron rich and are likely unstable. Many of the short-lived radioactive nuclei used in medicine are produced in fission reactions in nuclear reactors.