Suppose a ${}^{\mathbf{238}}\mathbf{U}$nucleus “swallows” a neutron and then decays not by fission but by beta-minus decay, in which it emits an electron and a neutrino. Which nuclide remains after this decay ${}^{\mathbf{239}}\mathit{P}\mathit{u}\mathbf{,}\mathbf{}{}^{\mathbf{238}}\mathit{N}\mathit{p}\mathbf{,}\mathbf{}{}^{\mathbf{239}}\mathbf{Np}\mathbf{,}\mathbf{}\mathbf{or}\mathbf{}{}^{\mathbf{238}}\mathbf{Pa}$?

The mass number of the atom remains the same, and the atomic number of the atom increases by one unit due to the decay of one beta particle from the atom.

The mass number of an atom is equal to the sum of the number of protons and neutrons.

The mass number of Uranium is 238, so the sum of protons and neutrons from uranium is 238, but it swallows one neutron, so the sum of protons and neutrons for new nuclei becomes 239.

The Uranium is also decaying as a beta particle, so the atomic number of new nuclei increases by one unit. The atomic number of Uranium is 92, so the atomic number of new nuclei will be 93.

The mass number for the new element is 239, and the atomic number of the new element is 93. The element for the corresponding atomic number 93 is Neptune (Np), but the mass number for Neptune is 239.

Therefore, the nuclide remains after decay is ${}^{\mathbf{239}}\mathbf{Np}$.