Question 26.6: The Conversion of Mass to Kinetic Energy in Uranium Fission ...
The Conversion of Mass to Kinetic Energy in Uranium Fission
GOAL Understand the production of energy from nuclear sources.
PROBLEM The fission, or splitting, of uranium was discovered in 1938 by Lise Meitner, who successfully interpreted some curious experimental results found by Otto Hahn as due to fission. (Hahn received the Nobel Prize.) The fission of ^{235}_{92}U begins with the absorption of a slow-moving neutron that produces an unstable nucleus of ^{236}U. The ^{236}U nucleus then quickly decays into two heavy fragments moving at high speed, as well as several neutrons. Most of the kinetic energy released in such a fission is carried off by the two large fragments. (a) For the typical fission process,
^{1}_{0}\mathrm{n}+^{235}_{92}\mathrm{U}\ \ \ \rightarrow\ \ ^{141}_{56}{Ra}+^{92}_{36}{Kr}+3_{0}^{1}\mathrm{n}
calculate the kinetic energy in MeV carried off by the fission fragments, neglecting the kinetic energy of the reactants. (b) What percentage of the initial energy is converted into kinetic energy? The atomic masses involved, given in atomic mass units, are
^{1}_{0}\mathrm{n}=1.008\ 665\mathrm{\,u}\qquad{^{235}_{92}U}=235.043\ 923\mathrm{\,u}
\ _{\mathrm{{56}}}^{\mathrm{{141}}}\mathrm{{Ba}}=\mathrm{{l40.903~496~u}}\qquad{^{92}_{36}{Kr}}=9\mathrm{{l.907~936~u}}
STRATEGY This problem is an application of the conservation of relativistic energy. Write the conservation law as a sum of kinetic energy and rest energy, and solve for the final kinetic energy.
(a) Calculate the final kinetic energy for the given process.
Apply the conservation of relativistic energy equation, assuming K E_{\mathrm{initial}}= 0:
(K E+m c^{2})_{\mathrm{initial}}=(K E+m c^{2})_{\mathrm{final}}
0\,+\,m_{\mathrm{n}}c^{2}\,+\,m_{\mathrm{U}}c^{2}=\,m_{\mathrm{B}a}c^{2}\,+\,m_{\mathrm{Kr}}c^{2}+\,3\,m_{\mathrm{n}}c^{2}+\,K E_{\mathrm{final}}
Solve for K E_{\mathrm{final}} and substitute, converting to MeV in the last step:
K E_{\mathrm{final}}=\left[(m_{\mathrm{n}}+m_{\mathrm{U}})\,-\,(m_{\mathrm{Ba}}+m_{\mathrm{Kr}}+3m_{\mathrm{n}})\right]c^{2}
K E_{\mathrm{final}}=(1.008~665~u+235.043~923~u)c^{2}
-\ [140.903\ 496\ \mathrm{u}+91.907\ 936\ \mathrm{u}+3(1.008\ 665\ \mathrm{u})]c^{2}
= (0.215 161 u)(931.494 MeV/u·c²)(c²)
= 200.421 MeV
(b) What percentage of the initial energy is converted into kinetic energy?
Compute the total energy, which is the initial energy:
E_{\mathrm{initial}}=0+\,m_{\mathrm{n}}c^{2}+\,m_{\mathrm{U}}c^{2}
= (1.008 665 u + 235.043 923 u)c²
= (236.052 59 u)(931.494 MeV/u·c²)(c²)
=2.198\ 82\times10^{5}\,\mathrm{MeV}
Divide the kinetic energy by the total energy and multiply by 100%:
\frac{200.421~\mathrm{MeV}}{2.198~82~\times~10^{5}\,\mathrm{MeV}}\times100% = 9.115\times10^{-{{2}}}%
REMARKS This calculation shows that nuclear reactions liberate only about one-tenth of 1% of the rest energy of the constituent particles. Some fusion reactions result in a percent yield several times as large.