Question

The nuclide${\mathbf{}}^{\mathbf{238}}\mathbf{Np}$ requires4.2 MeVfor fission. To remove a neutron from this nuclide requires an energy expenditure of 5.0 MeV. Isfissionable by thermal neutrons?

Short answer

The ${}^{237}\mathrm{Np}$is fissionable by thermal neutrons.

Step-by-step solution

A concept and identification of given data:

The energy required for the fission of a nuclide varies with the number of the nucleon in the nuclide or the mass number of that nuclide.

The energy required for the fission of ${}^{238}\mathrm{Np}$is ${\mathrm{E}}_1=4.2\mathrm{MeV}$.

The energy required for the fission of ${}^{237}\mathrm{Np}$is ${\mathrm{E}}_2=5\mathrm{MeV}$.

Whether Np-237 is fissionable by thermal neutrons:

The mass number of ${}^{238}\mathrm{Np}$is ${}^{237}\mathrm{Np}$and energy required for fission is 4.2 MeV. The mass number of ${}^{237}\mathrm{Np}$ is 237 and energy required for fission is 5 Me V which is more than the energy required for fission of ${}^{238}\mathrm{Np}$so the ${}^{237}\mathrm{Np}$is fissionable by thermal neutrons.

Hence, the ${}^{237}\mathrm{Np}$is fissionable by thermal neutrons.