Consider the fission reaction in which U-235 is bombarded by neutrons. The products of the bombardment are rubidium-89, cerium-144, beta particles, and more neutrons. (a) Write a balanced nuclear equation for the bombardment. (b) Calculate \(\Delta E\) when one gram of U-235 undergoes fission. (c) The detonation of TNT, an explosive, evolves \(2.76 \mathrm{~kJ} / \mathrm{g}\). How many kilograms of TNT are required to produce the same amount of energy as one milligram of \(\mathrm{U}-235\) ?

For a nuclear reaction, the number of protons and neutrons must be equal on both sides. We have U-235 (number of protons Z = 92, number of neutrons N = 143) bombarded by a neutron. The products are Rubidium-89 (Z = 37, N = 52), Cerium-144 (Z = 58, N = 86), a few beta particles, and some neutrons. Let's first balance the protons and neutrons: $$_{0}^{1}\mathrm{n} + _{92}^{235}\mathrm{U}\rightarrow _{37}^{89}\mathrm{Rb} + _{58}^{144}\mathrm{Ce} +x_\beta^{-1}\mathrm{e} +y_{0}^{1}\mathrm{n}$$ The total number of protons on both sides: 92 = 37 + 58 The total number of neutrons on both sides: 143 + 1 = 52 + 86 + y This gives us y = 6 for the number of neutrons and x = 3 for the number of beta particles: $$_{0}^{1}\mathrm{n} + _{92}^{235}\mathrm{U}\rightarrow _{37}^{89}\mathrm{Rb} + _{58}^{144}\mathrm{Ce} +3_\beta^{-1}\mathrm{e} +6_{0}^{1}\mathrm{n}$$

To calculate ΔE, we need first to find the binding energy of the initial and final products and then subtract them to find the energy difference. The mass of U-235 is 235.043 amu, and that of a neutron is 1.0087 amu. Thus, the mass of the reactants is: $$_i\mathrm{M}=235.043\,\mathrm{amu}+1.0087\,\mathrm{amu}=236.0517\,\mathrm{amu}$$ Since 1 amu = 931.5 MeV, the mass-energy of reactants is: $$_i\mathrm{E}=236.0517\,\mathrm{amu}\times931.5\,\frac{\mathrm{MeV}}{\mathrm{amu}}=219793.776\,\mathrm{MeV}$$ Now, we need to find the mass-energy of the products: Rubidium-89: mass = 88.919 amu Cerium-144: mass = 143.912 amu Neutron: mass = 1.0087 amu So the mass of products is $$_f\mathrm{M}=88.919\,\mathrm{amu}+143.912\,\mathrm{amu}+6\times1.0087\,\mathrm{amu}=234.8702\,\mathrm{amu}$$ The mass-energy of products is: $$_f\mathrm{E}=234.8702\,\mathrm{amu}\times931.5\,\frac{\mathrm{MeV}}{\mathrm{amu}}=218677.198\,\mathrm{MeV}$$ To find ΔE, we subtract the energy of the products from that of the reactants: $$\Delta E=_i\mathrm{E}-_f\mathrm{E}=219793.776\,\mathrm{MeV}-218677.198\,\mathrm{MeV}=1116.578\,\mathrm{MeV}$$ To find the energy released when _1_ gram of U-235 undergoes fission, we need to convert the MeV into Joules and multiply by the number of atoms in _1_ gram: $$\Delta E=1116.578\,\mathrm{MeV}\times\frac{1.602\times10^{-13}\,\mathrm{J}}{1\,\mathrm{MeV}}=1.79\times10^{-10}\,\mathrm{J}$$ For 1 gram of U-235: $$1\,\mathrm{g}\times\frac{1\,\mathrm{mol}}{235.043\,\mathrm{g}}\times\frac{6.022\times10^{23}}{1\,\mathrm{mol}}=2.563\times10^{21}\,\mathrm{atoms}$$ So, for 1 gram of U-235 undergoing fission: $$_b\Delta E=1.79\times10^{-10}\,\mathrm{J/atom}\times2.563\times10^{21}\,\mathrm{atoms}=4.59\times10^{11}\,\mathrm{J}$$

First, let's calculate the energy released by 1 mg of U-235: $$_c\Delta E=_b\Delta E\times\frac{1\,\mathrm{mg}}{1000\,\mathrm{mg}}=4.59\times10^{11}\,\mathrm{J}\times\frac{1}{1000}=4.59\times10^8\,\mathrm{J}$$ Now, calculate the mass of TNT needed to produce this amount of energy: $$\mathrm{Mass}\,\mathrm{of}\,\mathrm{TNT}=\frac{4.59\times10^{8}\,\mathrm{J}}{2.76\times10^3\,\mathrm{J/g}}=166181.159\,\mathrm{g}=166.181\, \mathrm{kg}$$ So, 166.181 kg of TNT is required to produce the same amount of energy as 1 milligram of U-235.