The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .\) (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{amu} ;{ }^{141} \mathrm{Ba}\) nuclear mass, 140.8833 amu; \({ }^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu \()\) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in 1.0 day?

To calculate the loss of mass from the Sun, we need to convert the energy falling on Earth to an equivalent mass using Einstein's mass-energy equivalence formula: \(E = mc^2\). We are given the energy per minute, so need to first calculate the energy per day. Energy per day: \(1.07 \times 10^{16} \mathrm{kJ/min} \times 1440 \mathrm{min/day}\) Now, we'll rearrange the formula to solve for mass: \(m = \frac{E}{c^2}\), where \(c = 3.0 \times 10^8 \;\text{m/s}\) is the speed of light.

To find the mass of uranium-235 required to produce 0.10% of the solar energy, we first need to determine the energy released by one uranium-235 nucleus undergoing fission. We are given the nuclear masses and the fission reaction. The energy released can be determined by calculating the mass defect. Mass defect = (Reactants mass - Products mass) = (mass of uranium-235 nucleus - (mass of Ba nucleus + mass of Kr nucleus)) \$Delta E\ = \Delta m \cdot c^2\$ Now, we can determine how many uranium-235 nuclei are required to produce 0.10% of the solar energy: Number of uranium-235 nuclei = \(\frac{0.001 \cdot \text{Energy per day}}{\Delta E}\) Finally, we can calculate the mass of uranium-235 required by multiplying the number of uranium-235 nuclei by the mass of one uranium-235 nucleus.