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, \(234.9935 \mathrm{u} ;{ }^{141} \mathrm{Ba}\) nuclear mass, $140.8833 \mathrm{u} ;{ }^{92} \mathrm{Kr}\( nuclear mass, \)91.9021 \mathrm{u}$ ) 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?

Multiply the energy per minute by the number of minutes in a day to get the total solar energy falling on Earth in a day: \[Energy_{day} = (1.07 \times 10^{16} \mathrm{~kJ/min}) \times (24 ~(hours) \times 60 ~(min))\] \[Energy_{day} = 1.54 \times 10^{19} \mathrm{~kJ}\]

Using Einstein's equation, \(E=mc^2\), we will find the mass loss of the Sun as: \[m = \frac{E}{c^2}\] Here, \(E = 1.54 \times 10^{19} \mathrm{~kJ}\) (Convert it to Joules: \(1.54 \times 10^{19} \mathrm{~kJ} \times 1000\) J/1 kJ = \(1.54 \times 10^{22} \mathrm{~J}\)) \(c = 3 \times 10^8 \mathrm{~m/s}\) Calculate the mass loss: \[m = \frac{1.54 \times 10^{22} \mathrm{~J}}{(3 \times 10^8 \mathrm{~m/s})^2}\] \[m \approx 1.71 \times 10^6 \mathrm{~kg}\]

For the given nuclear reaction, mass of \({ }^{235} \mathrm{U}\), \({ }^{141} \mathrm{Ba}\), and \({ }^{92} \mathrm{Kr}\) are given. Also, we know that the mass of a neutron is approximately 1u. Therefore, we can find the energy released in the reaction by using the mass-energy equivalence: Energy released = \((m_{U} - m_{Ba} - m_{Kr} - 4(m_{n}))c^2\) Using the values given, \(m_{U} = 234.9935, m_{Ba} = 140.8833\) \(m_{Kr} = 91.9021, m_{n} \approx 1\) Calculate the energy released per reaction: Energy released = \((234.9935 - 140.8833 - 91.9021 - 4(1)) \times 931.5(\mathrm{MeV})\) Energy released = \(200.2951 \times 931.5(\mathrm{MeV}) \approx 186.8(\mathrm{MeV})\)

First, let's find out 0.10% of the solar energy falling on Earth: \[0.1 \% \mathrm{~of~Solar~Energy} = \frac{0.1}{100} \times 1.54 \times 10^{19} \mathrm{~kJ}\] \[= 1.54 \times 10^{17} \mathrm{~kJ}\] Now, to find how many reactions are required to produce this amount of energy, divide the total energy required by energy produced per reaction: \[\mathrm{Number~of~Reactions} = \frac{1.54 \times 10^{17} \mathrm{~kJ}}{186.8\mathrm{~MeV}}\] To convert the numerator to MeV, multiply by \(1.60218\times 10^{-10}\): \[\mathrm{Number~of~Reactions} = \frac{1.54 \times 10^{17} \mathrm{~kJ} \times 1.60218\times 10^{-10} \mathrm{MeV/kJ}}{186.8\mathrm{~MeV}}\] \[\mathrm{Number~of~Reactions} \approx 8.756\times 10^{23}\] Now, knowing the number of reactions, we can find the mass of uranium-235 required: Mass of Uranium-235 = Number of Reactions \( \times \frac{m_{U}}{N_{A}}\) Here, \(m_{U} = 234.9935\) and \(N_{A}\) is the Avogadro's constant, \(6.022 \times 10^{23}\). Calculate the mass of Uranium-235: \[Mass~of~Uranium-235 = (8.756\times 10^{23}) \times \frac{234.9935}{6.022 \times 10^{23}}\] \[Mass~of~Uranium-235 \approx 3.389 \times 10^3 \mathrm{~kg}\] Answer: (a) The loss of mass from the Sun due to solar energy falling on Earth's surface is approximately \(1.71 \times 10^6\) kg per day. (b) The mass of uranium-235 required to produce 0.10% of the solar energy that falls on Earth in a day is approximately \(3.389 \times 10^3\) kg.