Animated Solution for Physics - Atoms and Nuclei: In a nuclear reactor ${}^{235}\text{U}$ undergoes fission liberating $200\text{ MeV}$ of energy. The reactor has a $10\%$ efficiency and produces $1000\text{ MW}$ power. If the reactor is to function for $10\text{ yr}$, find the total mass of uranium required.

Imagine a power plant capable of generating $1000\text{ MW}$ of electricity continuously for an entire decade. That is an astronomical amount of energy, enough to power a massive city without a single interruption. But where does this energy come from? It comes from the splitting of tiny atomic nuclei—specifically, Uranium-235. In this problem, we are tasked with finding out exactly how much uranium is needed to fuel this colossal endeavor.

First, we need to understand the sheer scale of the energy required. The reactor produces a useful power output of $P_{\text{out}}=1000\text{ MW}$, which is equivalent to $10^9\text{ J/s}$.

It needs to sustain this output for $10\text{ years}$. Let's convert this time into seconds: $t=10\times 365\times 24\times 3600=3.1536\times 10^8\text{ s}$

The total useful energy delivered over this decade is simply power multiplied by time: $E_{\text{out}}=P_{\text{out}}\times t=10^9\times 3.1536\times 10^8=3.1536\times 10^{17}\text{ J}$

However, no machine is perfect. Our nuclear reactor operates at a modest efficiency of $10\%$. This means that for every $100\text{ Joules}$ of thermal energy generated by the splitting atoms, only $10\text{ Joules}$ are successfully converted into electrical power.

To find the total thermal energy that the uranium must produce, we divide our useful output by the efficiency: $E_{\text{in}}=\frac{E_{\text{out}}}{\eta }=\frac{3.1536\times 10^{17}}{0.1}=3.1536\times 10^{18}\text{ J}$

Now, let's zoom in to the atomic level. Every time a single ${}^{235}\text{U}$ nucleus undergoes fission, it releases $200\text{ MeV}$ of energy. To compare this with our total energy requirement, we must convert it into Joules: $E_f=200\times 10^6\times 1.6\times 10^{-19}=3.2\times 10^{-11}\text{ J}$

How many of these tiny atomic explosions are needed to reach our massive energy goal? We divide the total required energy by the energy per fission: $N=\frac{E_{\text{in}}}{E_f}=\frac{3.1536\times 10^{18}}{3.2\times 10^{-11}}=0.9855\times 10^{29}\text{ atoms}$

We have the number of atoms, but engineers don't order uranium by the atom—they order it by the kilogram. We use Avogadro's number to convert atoms into moles. Since we want our final answer in kilograms, we'll use the kilo-mole version of Avogadro's number ($N_A=6.02\times 10^{26}{\text{ kmol}}^{-1}$): $n=\frac{N}{N_A}=\frac{0.9855\times 10^{29}}{6.02\times 10^{26}}=163.7\text{ kmol}$

Finally, we multiply the number of kilo-moles by the molar mass of Uranium-235 ($235\text{ kg/kmol}$) to find the total mass: $m=n\times M=163.7\times 235=38469.5\text{ kg}$

Rounding to an appropriate number of significant figures, we get our final answer: $m\approx 3.847\times 10^4\text{ kg}$

It takes nearly $38.5\text{ tonnes}$ of pure Uranium-235 to keep this city lit for a decade!