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Found in: Page 1331

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# The nuclide${{\mathbf{}}}^{{\mathbf{238}}}{\mathbf{Np}}$ requires 4.2 MeV for fission. To remove a neutron from this nuclide requires an energy expenditure of 5.0 MeV. Is fissionable by thermal neutrons?

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

See the step by step solution

## Step 1: 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 ${\mathrm{E}}_{2}=5\mathrm{MeV}$ .

## Step 2: 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.

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