Question 43.10: The Energy Released in the Fission of ^235U Calculate the en...
The Energy Released in the Fission of { }^{\pmb{235}} \pmb{U}
Calculate the energy released when 1.00 kg of { }^{235} U fissions, taking the disintegration energy per event to be Q = 208 MeV.
Conceptualize Imagine a nucleus of { }^{235} U absorbing a neutron and then splitting into two smaller nuclei and several neutrons as in Figure 43.17.
Categorize The problem statement tells us to categorize this example as one involving an energy analysis of nuclear fission.
Analyze Because A = 235 for uranium, one mole of this isotope has a mass of M = 235 g.
Find the number of nuclei in our sample in terms of the number of moles n and Avogadro’s number, and then in terms of the sample mass m and the molar mass M of { }^{235} U :
N=n N_{A}=\frac{m}{M} N_{A}Find the total energy released when all nuclei undergo fission:
\begin{aligned}E & =N Q=\frac{m}{M} N_{A} Q=\frac{1.00 \times 10^3 g}{235 g/ \text{mol}}\left(6.02 \times 10^{23} \text{mol}^{-1}\right)(208 MeV) \\& =5.33 \times 10^{26} MeV\end{aligned}Finalize Convert this energy to kWh:
E=\left(5.33 \times 10^{26} MeV\right)\left(\frac{1.60 \times 10^{-13} J}{1 MeV}\right)\left(\frac{1 kWh}{3.60 \times 10^6 J}\right)=2.37 \times 10^7 kWhwhich, if released slowly, is enough energy to keep a 100-W lightbulb operating for 30 000 years! If the available fission energy in 1 kg of { }^{235} U were suddenly released, it would be equivalent to detonating about 20 000 tons of TNT.
