91Ó°ÊÓ

Energy is a measurable attribute that may be transferred from one thing to another in order for it to do work.

\(\begin{array}{c}m({}^{238}U) = 238.0507882\,{\rm{u}}\\m({}^{234}Th) = 234.043593\,{\rm{u}}\\m(\alpha ) = 4.00260325415\,{\rm{u}}\end{array}\)

As a result, the mass difference is,

\(\begin{array}{c}\Delta m = m({}^{238}U) - \{ [m({}^{234}Th)] + [m(\alpha )]\} \\ = \left( {238.0507882\,{\rm{u}}} \right) - \{ (234.043593\,{\rm{u}}) + (4.00260325415\,{\rm{u}})\} \\ = 0.0045919\,{\rm{u}}\end{array}\)

We can calculate the energy released during the alpha decay.

\(\begin{array}{c}E = \Delta m{c^2}\\E = (0.0045919\,{\rm{u}})(931.5\,{\rm{MeV/u}}{{\rm{c}}^2}) \times {c^2}\\ = 4.2774\,{\rm{MeV}}\end{array}\)

Therefore, the energy released is \({\rm{4}}{\rm{.2774MeV}}\).

\(\begin{array}{c}\frac{{\Delta m}}{{m({}^{238}U)}} = \frac{{(0.0045919\,{\rm{u}})}}{{\left( {238.0507882\,{\rm{u}}} \right)}}\\ = 1.92 \times {10^{ - 5}}\end{array}\)

Therefore, the fraction of mass destroyed is \(1.92 \times {10^{ - 5}}\).

(c) Since the half-life of\({}^{{\rm{238}}}{\rm{U}}\) is so long \(\left( {4.468 \times {{10}^9}\,{\rm{year}}} \right)\), just a few atoms will decay in a macroscopic sample of Uranium. As a result, the decrease in mass is barely noticeable.