Calculating Energy from a Kilogram of Fissionable Fuel

The total energy produced is the number of ${}^{235}\text{U}$ atoms times the given energy per ${}^{235}\text{U}$ fission. We should therefore find the number of ${}^{235}\text{U}$ atoms in 1.00 kg.

The number of ${}^{235}\text{U}$ atoms in 1.00 kg is Avogadro’s number times the number of moles. One mole of ${}^{235}\text{U}$ has a mass of 235.04 g; thus, there are $\left(\mathrm{1,000}\text{g}\right)/\left(\mathrm{235.04.00}\text{g}/\text{m}\text{o}\text{l}\right)=4.25\text{m}\text{o}\text{l}$ . The number of ${}^{235}\text{U}$ atoms is therefore

$$\left(4.25\text{mol}\right)\left(6.02\times {10}^{23}\text{U}/\text{mol}\right)=2.56\times {10}^{24}\text{atoms of}{}^{235}\text{U}.$$

So the total energy released is