Question

A thermal neutron (with approximately zero kinetic energy) is absorbed by a238Unucleus. How much energy is transferred from mass-energy to the resulting oscillation of the nucleus? Here are some atomic masses and neutron mass.

$$\begin{gathered} {}^{\mathbf{237}}\mathbf{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{237}\mathbf{.}\mathbf{048723}\mathbf{}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{237}}\mathbf{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{238}\mathbf{.}\mathbf{050782}\mathbf{}\mathbf{u} \\ {}^{\mathbf{237}}\mathbf{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{239}\mathbf{.}\mathbf{054287}\mathbf{}\mathbf{u}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{}^{\mathbf{237}}\mathbf{U}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{240}\mathbf{.}\mathbf{056585}\mathbf{}\mathbf{u} \\ \mathbf{n}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{008664}\mathbf{}\mathbf{u}\mathbf{} \end{gathered}$$

Short answer

The energy transferred from mass energy to oscillation of the nucleus is 4.8 MeV .

Step-by-step solution

A concept and identification of given data:

The difference in the mass of the reactant and product in the nuclear reaction is called the mass defect. This mass is converted into mass energy.

$$\begin{gathered} \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{U}-239\mathrm{is}{\mathrm{m}}_{239}=239.054287\mathrm{u} \\ \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{U}-239\mathrm{is}{\mathrm{m}}_{238}=238.050782\mathrm{u}. \\ \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{a}\mathrm{neutron}\mathrm{is}{\mathrm{m}}_{\mathrm{n}}=1.008664\mathrm{u}. \end{gathered}$$

Determination of energy transferred from mass energy to oscillation of nucleus:

The energy transferred from mass energy to oscillation of nucleus is given as:

$E=\left(m_{238}+m_n-m_{239}\right)c^2$

Here, c is the speed of light and its equivalent mass energy is 931.5 MeV/u.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{E}=\left[238.050782\mathrm{u}+1.008664\mathrm{u}-239.054287\mathrm{u}\right]\left(\frac{931.5\mathrm{MeV}}{\mathrm{u}}\right) \\ =\left[0.00519\mathrm{u}\right]\left(\frac{931.5\mathrm{MeV}}{\mathrm{u}}\right) \\ =4.8\mathrm{MeV} \end{gathered}$$

Hence, the energy transferred from mass energy to oscillation of nucleus is 4.8 MeV