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The Law of Radioactive Decay (HL) (HL IB Physics)

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Katie M

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Katie M

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The Law of Radioactive Decay

  • In radioactive decay, the number of undecayed nuclei falls very rapidly, without ever reaching zero
    • Such a model is known as exponential decay

  • The graph of number of undecayed nuclei against time has a very distinctive shape:

Exponential Decay Graph, downloadable AS & A Level Physics revision notes

Radioactive decay follows an exponential pattern. The graph shows three different isotopes each with a different rate of decay

  • The key features of this graph are:
    • The steeper the slope, the larger the decay constant λ (and vice versa)
    • The decay curves always start on the y-axis at the initial number of undecayed nuclei (N0)

Equations for Radioactive Decay

  • The number of undecayed nuclei N can be represented in exponential form by the equation:

N space equals space N subscript 0 space e to the power of negative lambda t end exponent

  • Where:
    • N0 = the initial number of undecayed nuclei (when t = 0)
    • N = number of undecayed nuclei at a certain time t
    • λ = decay constant (s-1)
    • t = time interval (s)
  • The number of nuclei can be substituted for other quantities.
  • For example, the activity A is directly proportional to N, so it can also be represented in exponential form by the equation:

A space equals space A subscript 0 space e to the power of negative lambda t end exponent

  • Where:
    • A = activity at a certain time t (Bq)
    • A0 = initial activity (Bq)
  • The received count rate C is related to the activity of the sample, hence it can also be represented in exponential form by the equation:

C space equals space C subscript 0 space e to the power of negative lambda t end exponent

  • Where:
    • C = count rate at a certain time t (counts per minute or cpm)
    • C0 = initial count rate (counts per minute or cpm)

Worked example

Strontium-90 decays with the emission of a β-particle to form Yttrium-90. The decay constant of strontium-90 is 0.025 year -1.

Determine the activity A of the sample after 5.0 years, expressing the answer as a fraction of the initial activity A0.

Answer:

Step 1: Write out the known quantities

  • Decay constant, λ = 0.025 year -1
  • Time interval, t = 5.0 years
  • Both quantities have the same unit, so there is no need for conversion

Step 2: Write the equation for activity in exponential form

A space equals space A subscript 0 space e to the power of negative lambda t end exponent

Step 3: Rearrange the equation for the ratio between A and A0

A over A subscript 0 space equals space e to the power of negative lambda t end exponent

Step 4: Calculate the ratio A/A0

A over A subscript 0 space equals space e to the power of negative open parentheses 0.025 cross times 5 close parentheses end exponent space equals space 0.88

  • Therefore, the activity of strontium-90 decreases by a factor of 0.88, or 12%, after 5 years

Worked example

A space probe uses a source containing 4.0 kg of plutonium-238.

Plutonium-238 is an alpha-emitter with a half-life of 87.7 years. Each alpha decay releases 5.5 MeV per emission. The space probe converts this into electrical energy with an efficiency of 32%.

The space probe can continue to operate as long as the power output is maintained at 0.4 kW or above.

Estimate the time, in years, the source is expected to supply power to the space probe.

Answer:

Step 1: List the known quantities

  • Mass of Pu-238 = 4.0 kg = 4000 g
  • Molar mass of Pu-238 = 238 g mol−1
  • Avogadro's constant, N subscript A = 6.02 × 1023 mol−1
  • Half-life of Pu-238 = 87.7 years
  • Energy released per alpha decay = 5.5 MeV
  • 1 electronvolt (eV) = 1.6 × 10−19 J
  • Efficiency = 32% = 0.32
  • Final power output, P = 0.4 kW = 400 W

Step 2: Calculate the initial number of nuclei present in the source

  • 238 g of plutonium-238 contains 6.02 × 1023 atoms (Avogadro's number), so in 4 kg:

Number of nuclei: N space equals space fraction numerator m a s s space cross times space N subscript A over denominator m o l a r space m a s s end fraction

Initial number of nuclei: N subscript 0 space equals space fraction numerator 4000 space cross times space open parentheses 6.02 cross times 10 to the power of 23 close parentheses over denominator 238 end fraction space equals space 1.012 cross times 10 to the power of 25 nuclei

Step 3: Calculate the initial activity of the source

Decay constant:  lambda space equals space fraction numerator ln space 2 over denominator t subscript 1 divided by 2 end subscript end fraction

Activity: A space equals space lambda N

  • Combining these gives:

Initial activity: A subscript 0 space equals space fraction numerator N subscript 0 space ln space 2 over denominator t subscript 1 divided by 2 end subscript end fraction

A subscript 0 space equals space fraction numerator open parentheses 1.012 cross times 10 to the power of 25 close parentheses space cross times space ln space 2 over denominator 87.7 space cross times space open parentheses 24 cross times 60 cross times 60 cross times 365 close parentheses end fraction space equals space 2.54 cross times 10 to the power of 15Bq

Step 4: Calculate the initial power output of the source

Power output: P space equals space fraction numerator increment E over denominator increment t end fraction

Energy released per decay: E space equals space open parentheses 5.5 cross times 10 to the power of 6 close parentheses cross times open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses space equals space 8.8 cross times 10 to the power of negative 13 end exponent J

  • Activity represents the decays per second, so:

Initial power output: P subscript 0 space equals space A subscript 0 E

P subscript 0 space equals space open parentheses 2.5 cross times 10 to the power of 15 close parentheses space cross times space open parentheses 8.8 cross times 10 to the power of negative 13 end exponent close parentheses space equals space 2200 W

  • The electrical power transferred to the probe is:

P subscript 0 space equals space 2200 space cross times space 0.32 space equals space 704 W

Step 5: Use the exponential decay equation to calculate the time of operation

  • The power available is proportional to the activity of the isotope, so: 

Exponential decay of power: P space equals space P subscript 0 space e to the power of negative lambda t end exponent

fraction numerator P over denominator P subscript 0 space end fraction space equals space e to the power of negative lambda t end exponent

ln open parentheses fraction numerator P over denominator P subscript 0 space end fraction close parentheses space equals space minus lambda t

t space equals space minus 1 over lambda space ln open parentheses P over P subscript 0 close parentheses space equals space minus fraction numerator t subscript 1 divided by 2 end subscript over denominator ln space 2 end fraction space ln open parentheses P over P subscript 0 close parentheses

t space equals space minus fraction numerator 87.7 over denominator ln space 2 end fraction space ln open parentheses 400 over 704 close parentheses space equals space 71.5 years

  • Therefore, the source is expected to supply power to the space probe for 71.5 years

Examiner Tip

The symbol e is used to represent the exponential constant and is approximately equal to e = 2.718

Make sure you are comfortable using the exponential function on your calculator, it is the button labelled e to the power of x

The inverse function of e to the power of x is the natural logarithmic function, ln space y

The rules for exponential functions are the same as the rules for logarithmic functions, so, if y space equals space e to the power of x, then x space equals space ln space y

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    Katie M

    Author: Katie M

    Expertise: Physics

    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.