Energy Released in Fission Reactions (DP IB Physics: SL): Revision Note

When a uranium-235 nucleus absorbs a slow-moving neutron and undergoes a fission reaction, one possible pair of fission fragments is technetium-112 and indium-122.

The equation for this process, and the binding energy per nucleon for each isotope, are shown below.

${}_{92}{}^{235}\mathrm{U}+{}_0{}^1\mathrm{n}\to {}_{43}{}^{112}\mathrm{Tc}+{}_{49}{}^{122}\mathrm{In}+\hspace{0.17em}2{}_0{}^1\mathrm{n}$

(a) Calculate the energy released per fission of uranium-235, in MeV.

(b) Determine the mass of uranium-235 required per day to run a 500 MW power plant at 35% efficiency.

(c) The specific energy of coal is approximately 35 MJ kg⁻¹. 

For the same power plant, estimate the ratio 

$\frac{\mathrm{mass}\mathrm{of}\mathrm{coal}\mathrm{required}\mathrm{per}\mathrm{day}}{\mathrm{mass}\mathrm{of}{}^{235}\mathrm{U}\mathrm{required}\mathrm{per}\mathrm{day}}$

Answer:

(a)  Energy released per fission of uranium-235

Step 1: Determine the binding energies of the nuclei before and after the reaction

• Binding energy is equal to binding energy per nucleon × mass number

• Binding energy before $\left({}_{}{}^{235}\mathrm{U}\right)$ = 235 × 7.59 = 1784 MeV

• Binding energy after $\left({}_{}{}^{112}\mathrm{Tc}+{}_{}{}^{122}\mathrm{In}\right)$ = (112 × 8.36) + (122 × 8.51) = 1975 MeV

Step 2: Find the difference to obtain the energy released per fission reaction

• Therefore, the energy released per fission = 1975 – 1784 = 191 MeV

(b)  Mass of uranium-235 required per day

Step 1: List the known quantities

• Avogadro's number, N_A = 6.02 × 10²³ mol⁻¹

• Molar mass of U-235, m_r = 235 g mol⁻¹

• Power output, P_out = 500 MW = 500 × 10⁶ J s⁻¹

• Efficiency, e = 35% = 0.35

• Time, t = 1 day = 60 × 60 × 24 = 86 400 s

Step 2: Determine the number of nuclei in 1 kg of U-235

• There are N_A (Avogadro’s number) atoms in 1 mol of U-235, which is equal to a mass of 235 g

number of nuclei = $\frac{\mathrm{mass}\left(\mathrm{g}\right)\times N_A\left({\mathrm{mol}}^{-1}\right)}{m_r\left(\mathrm{g}{\mathrm{mol}}^{-1}\right)}$

• A mass of 1 kg (1000 g) of U-235 contains $\frac{1000\times \left(6.02\times {10}^{23}\right)}{235}$ = 2.562 × 10²⁴ atoms kg⁻¹

Step 3: Determine the specific energy of U-235

• Specific energy of U-235 = total amount of energy released by 1 kg of U-235

• Specific energy of U-235 = (number of atoms per kg) × (energy released per atom) = energy released per kg

• Energy released per atom of U-235 = 191 MeV

• Therefore, specific energy of U-235 = (2.562 × 10²⁴) × 191 = 4.893 × 10²⁶ MeV kg⁻¹

• To convert 1 MeV = 10⁶ × (1.6 × 10⁻¹⁹) J

• Specific energy of U-235 = (4.893 × 10²⁶) × 10⁶ × (1.6 × 10⁻¹⁹) = 7.83 × 10¹³ J kg⁻¹

Step 4: Use the relationship between power, energy and efficiency to determine the mass

• The input power required is:

efficiency & power:  $e=\frac{P_{out}}{P_{in}}\Rightarrow P_{in}=\frac{P_{out}}{e}$

input power:  $P_{in}=\frac{500}{0.35}=1429\mathrm{MW}$

$P_{in}=\frac{E_{in}}{t}=1429\times {10}^6\mathrm{J}\hspace{0.17em}{\mathrm{s}}^{-1}$

• Therefore, the mass of U-235 required in a day is:

$\mathrm{mass}\mathrm{of}{}^{235}\mathrm{U}\left(\mathrm{kg}{\mathrm{s}}^{-1}\right)=\frac{E_{in}\left(\mathrm{J}\right)}{1\mathrm{s}}\times \frac{1\mathrm{kg}}{\mathrm{specific}\mathrm{energy}\mathrm{of}{}^{235}\mathrm{U}\left(\mathrm{J}\right)}$

mass of U-235 (per second) =$\frac{1429\times {10}^6}{1}\times \frac{1}{7.83\times {10}^{13}}=1.82\times {10}^{-5}\mathrm{kg}{\mathrm{s}}^{-1}$

mass of U-235 (per day) = (1.82 × 10⁻⁵) × 86 400 = 1.58 kg

• Therefore, 1.58 kg of uranium-235 is required per day to run a 500 MW power plant at 35% efficiency 

(c)  Ratio of the masses of coal and U-235

• Since specific energy $\propto \frac{1}{\mathrm{mass}}$

$\frac{\mathrm{specific}\mathrm{energy}\mathrm{of}{}^{235}\mathrm{U}}{\mathrm{specific}\mathrm{energy}\mathrm{of}\mathrm{coal}}\propto \frac{\mathrm{mass}\mathrm{of}\mathrm{coal}\mathrm{required}\mathrm{per}\mathrm{day}}{\mathrm{mass}\mathrm{of}{}^{235}\mathrm{U}\mathrm{required}\mathrm{per}\mathrm{day}}$

• Where the energy density of coal = 35 MJ kg⁻¹

$\frac{\mathrm{mass}\mathrm{of}\mathrm{coal}\mathrm{required}\mathrm{per}\mathrm{day}}{\mathrm{mass}\mathrm{of}{}^{235}\mathrm{U}\mathrm{required}\mathrm{per}\mathrm{day}}=\frac{7.83\times {10}^{13}}{35\times {10}^6}=2.24\times {10}^6$

• Over 2 million times (~3.5 × 10⁶ kg) more coal is required than uranium-235 to achieve the same power output in a day (or second, or month or year)

If you need to brush up on binding energy calculations, take a look at the Mass Defect & Nuclear Binding Energy revision notes.