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

Katie M

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Katie M

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DP Physics: HLDP Physics: SL

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

density space open parentheses kg space straight m to the power of negative 3 end exponent close parentheses space equals space fraction numerator energy space density space open parentheses straight J space straight m to the power of negative 3 end exponent close parentheses over denominator specific space energy space open parentheses straight J space kg to the power of negative 1 end exponent close parentheses end fraction

number space of space nuclei space equals space fraction numerator mass space open parentheses straight g close parentheses space cross times space Avogadro apostrophe straight s space number space N subscript A space open parentheses mol to the power of negative 1 end exponent close parentheses over denominator molar space mass space open parentheses straight g space mol to the power of negative 1 end exponent close parentheses end fraction

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.

straight U presubscript 92 presuperscript 235 space plus space straight n presubscript 0 presuperscript 1 space rightwards arrow space Tc presubscript 43 presuperscript 112 space plus space In presubscript 49 presuperscript 122 space plus thin space 2 straight n presubscript 0 presuperscript 1

nucleus

binding energy per nucleon / MeV

straight U presubscript 92 presuperscript 235

7.59

Tc presubscript 43 presuperscript 112

8.36

In presubscript 49 presuperscript 122

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 

fraction numerator mass space of space coal space required space per space day over denominator mass space of space to the power of 235 straight U space required space per space day end fraction

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 open parentheses straight U presubscript blank presuperscript 235 close parentheses = 235 × 7.59 = 1784 MeV

  • Binding energy after open parentheses Tc presubscript blank presuperscript 112 space plus space In presubscript blank presuperscript 122 close parentheses = (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, NA = 6.02 × 1023 mol−1

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

  • Power output, Pout = 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 NA (Avogadro’s number) atoms in 1 mol of U-235, which is equal to a mass of 235 g

number of nuclei = fraction numerator mass space open parentheses straight g close parentheses space cross times space N subscript A space open parentheses mol to the power of negative 1 end exponent close parentheses over denominator m subscript r space open parentheses straight g space mol to the power of negative 1 end exponent close parentheses end fraction

  • A mass of 1 kg (1000 g) of U-235 contains fraction numerator 1000 space cross times space open parentheses 6.02 cross times 10 to the power of 23 close parentheses over denominator 235 end fraction = 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 space equals space P subscript o u t end subscript over P subscript i n end subscript space space space space space rightwards double arrow space space space space space P subscript i n end subscript space equals space P subscript o u t end subscript over e

input power:  P subscript i n end subscript space equals space fraction numerator 500 over denominator 0.35 end fraction space equals space 1429 space MW

P subscript i n end subscript space equals space E subscript i n end subscript over t space equals space 1429 space cross times 10 to the power of 6 space straight J thin space straight s to the power of negative 1 end exponent

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

mass space of space to the power of 235 straight U space open parentheses kg space straight s to the power of negative 1 end exponent close parentheses space equals space fraction numerator E subscript i n end subscript space open parentheses straight J close parentheses over denominator 1 space straight s end fraction cross times fraction numerator 1 space kg over denominator specific space energy space of space to the power of 235 straight U space open parentheses straight J close parentheses end fraction

mass of U-235 (per second) =fraction numerator 1429 cross times 10 to the power of 6 over denominator 1 end fraction cross times fraction numerator 1 over denominator 7.83 cross times 10 to the power of 13 end fraction space equals space 1.82 space cross times space 10 to the power of negative 5 end exponent space kg space straight s to the power of negative 1 end exponent

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 proportional to space 1 over mass

fraction numerator specific space energy space of space to the power of 235 straight U over denominator specific space energy space of space coal end fraction space proportional to space fraction numerator mass space of space coal space required space per space day over denominator mass space of space to the power of 235 straight U space required space per space day end fraction

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

fraction numerator mass space of space coal space required space per space day over denominator mass space of space to the power of 235 straight U space required space per space day end fraction space equals space fraction numerator 7.83 cross times 10 to the power of 13 over denominator 35 cross times 10 to the power of 6 end fraction space equals space 2.24 space cross times space 10 to the power of 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 Tips and Tricks

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

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Katie M

Author: Katie M

Expertise: Physics Content Creator

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.