Question

Breeder reactors are used to convert the nonfissionable nuclide \({ }_{92}^{238} \mathrm{U}\) to a fissionable product. Neutron capture of the \({ }_{92}^{238} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?

Short answer

The final fissionable product after two successive beta decays of \({}_{92}^{238}\mathrm{U}\) is Plutonium-239 (\({}_{94}^{239}\mathrm{Pu}\)).

Step-by-step solution

Neutron Capture by U-238

First, we'll identify the new nuclide formed after neutron capture. When a neutron is captured by an atom, the mass number increases by one, and the atomic number remains the same. Thus, after capturing a neutron, U-238 becomes: \({}_{92}^{239}\mathrm{U}\)

First Beta Decay

Now, we'll identify the nuclide after the first beta decay. During a beta decay, a neutron in the nucleus is converted into a proton, and an electron (beta particle) is emitted. This causes the atomic number to increase by 1, while the mass number remains the same. So after the first beta decay, the nuclide becomes: \({}_{93}^{239}\mathrm{Np}\) (Neptunium-239)

Second Beta Decay

Finally, we'll identify the final fissionable product after the second beta decay. Similarly, as the previous step, the atomic number increases by 1, and the mass number remains the same. After the second beta decay, the nuclide becomes: \({}_{94}^{239}\mathrm{Pu}\) (Plutonium-239) The final fissionable product after two successive beta decays of U-238 is Plutonium-239.