Energy Released in Fission Reactions

When a large (parent) nucleus, such as uranium-235, undergoes a fission reaction, the daughter nuclei produced as a result will have a higher binding energy per nucleon than the parent nucleus
As a result of the mass defect between the parent nucleus and the daughter nuclei, energy is released

5-4-2-energy-released-in-fission-reactions-graph

Energy can be extracted from fission reactions due to the mass defect between parent and daughter nuclei

Nuclear fission is well-regarded as having the fuel source with the highest energy density of any fuel that is currently available to us (until fusion reactions become feasible)

Examples of Common Fuels: Energy Density and Specific Energy Table

8-1-1-energy-comparison-table_sl-physics-rn

Calculations involving energy released in fission reactions often require the use of equations found in an array of previous topics, such as

$d e n s i t y (k g m^{- 3}) = \frac{e n e r g y d e n s i t y (J m^{- 3})}{s p e c i f i c e n e r g y (J k g^{- 1})}$

$n o . o f a t o m s = \frac{m a s s (g) \times A v o g a d r o' s n u m b e r N_{A} (m o l^{- 1})}{m o l a r m a s s (g m o l^{- 1})}$

Worked example

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.

${}_{92}^{235}U + {}_{0}^{1}n \to {}_{43}^{112}{Tc} + {}_{49}^{122}{In} + 2 {}_{0}^{1}n$

nucleus	binding energy per nucleon / MeV
${}_{92}^{235}U$	7.59
${}_{43}^{112}{Tc}$	8.36
${}_{49}^{122}{In}$	8.51

(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

$\frac{m a s s o f c o a l r e q u i r e d p e r d a y}{m a s s o f^{235} U r e q u i r e d p e r d a y}$

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 $({}^{235}U)$ = 235 × 7.59 = 1784 MeV
Binding energy after $({}^{112}{Tc} + {}^{122}{In})$ = (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 atoms 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 atoms = $\frac{m a s s (g) \times N_{A} (m o l^{- 1})}{m_{r} (g m o l^{- 1})}$

A mass of 1 kg (1000 g) of U-235 contains $\frac{1000 \times (6.02 \times 10^{23})}{235}$ = 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: $e = \frac{P_{o u t}}{P_{i n}} \Rightarrow P_{i n} = \frac{P_{o u t}}{e}$

input power: $P_{i n} = \frac{500}{0.35} = 1429 MW$

input power: $P_{i n} = \frac{E_{i n}}{t} = 1429 \times 10^{6} J s^{- 1}$

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

$m a s s o f^{235} U (k g s^{- 1}) = \frac{E_{i n} (J)}{1 s} \times \frac{1 k g}{s p e c i f i c e n e r g y o f^{235} U (J)}$

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

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 $\propto \frac{1}{m a s s}$

$\frac{s p e c i f i c e n e r g y o f^{235} U}{s p e c i f i c e n e r g y o f c o a l} \propto \frac{m a s s o f c o a l r e q u i r e d p e r d a y}{m a s s o f^{235} U r e q u i r e d p e r d a y}$

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

$\frac{m a s s o f c o a l r e q u i r e d p e r d a y}{m a s s o f^{235} U r e q u i r e d p e r d a y} = \frac{7.83 \times 10^{13}}{35 \times 10^{6}} = 2.24 \times 10^{6}$

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)

Examiner Tip

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

Energy Released in Fission Reactions (HL IB Physics)

Revision Note