Question

The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius R=7.4\times10^{-15} m. In nuclear fission, t

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The unstable nucleus uranium$$-236$$ can be regarded as auniformly charged sphere of charge $$Q=+92e$$ and radius $$\displaystyle{R}={7.4}\times{10}^{{-{15}}}$$ m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium$$-236$$ nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945.
A. Find the radii of the two "daughter" nuclei of charge$$+46e.$$
B. In a simple model for the fission process, immediatelyafter the uranium$$-236$$ nucleus has undergone fission the "daughter"nuclei are at rest and just touching. Calculate the kineticenergy that each of the "daughter" nuclei will have when they arevery far apart.
C. In this model the sum of the kinetic energies of the two"daughter" nuclei is the energy released by the fission of oneuranium$$-236$$ nucleus. Calculate the energy released by thefission of 10.0 kg of uranium$$-236$$. The atomic mass ofuranium$$-236$$ is 236 u, where 1 $$u = 1$$ atomic mass unit $$\displaystyle={1.66}\times{10}^{{-{27}}}$$ kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases $$4.18 \times 10^{12}$$ J when itexplodes).

2021-04-27
a.) The two daughter nuclei have half the volume of the originaluranium nucleus, so their radii are smaller by a factor of $$\displaystyle\sqrt{{{3}}}{\left\lbrace{2}\right\rbrace}$$
$$\displaystyle{r}={\frac{{{7.4}\times{10}^{{-{15}}}}}{{\sqrt{{{3}}}{\left\lbrace{2}\right\rbrace}}}}={5.9}\times{10}^{{-{15}}}$$ m
b) $$\displaystyle{U}={\frac{{{k}{\left({46}{e}\right)}^{{2}}}}{{{2}{r}}}}={\frac{{{k}{\left({46}\right)}^{{2}}{\left({1.6}\times{10}^{{-{19}}}\right)}^{{2}}}}{{{1.17}\times{10}^{{-{14}}}}}}={4.14}\times{10}^{{-{11}}}$$ J
Each daughter has half of the potential energy turn into itskinetic energy when far from each other, so
$$\displaystyle{K}={\frac{{{U}}}{{{2}}}}={\frac{{{4.14}\times{10}^{{-{11}}}}}{{{2}}}}={2.07}\times{10}^{{-{11}}}$$ J
c) If we have 10.0 kg of uranium, then the number of nucleiis
$$\displaystyle{n}={\frac{{{10}}}{{{236}{u}{\left({1.66}\times{10}^{{-{27}}}\right)}}}}={2.55}\times{10}^{{{25}}}$$ nuclei
And each releases energy:
$$\displaystyle{E}={n}{U}={\left({2.55}\times{10}^{{{25}}}\right)}{\left({4.14}\times{10}^{{-{11}}}\right)}={1.06}\times{10}^{{{15}}}$$ J
$$\displaystyle={253}$$ kilotons of TNT
d.) We could call an atomic bomb an “electric” bomb sincethe electric potential energy provides the kinetic energy of the particles