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Decay Constant & Half-Life (HL) (HL IB Physics)

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Katie M

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Katie M

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Decay Constant & Half-Life

  • Since radioactive decay is spontaneous and random, it is useful to consider the average number of nuclei that are expected to decay per unit time
    • This is known as the average decay rate
  • As a result, each radioactive element can be assigned a decay constant
  • The decay constant λ is defined as:

 The probability that an individual nucleus will decay per unit of time

  • When a sample is highly radioactive, this means the number of decays per unit time is very high
    • This suggests it has a high level of activity
  • Activity, or the number of decays per unit time can be calculated using:

A space equals space fraction numerator increment N over denominator increment t end fraction space equals space minus lambda N

  • Where:
    • A = activity of the sample (Bq)
    • ΔN = number of decayed nuclei
    • Δt = time interval (s)
    • λ = decay constant (s-1)
    • N = number of nuclei remaining in a sample
  • The activity of a sample is measured in becquerels (Bq)
    • An activity of 1 Bq is equal to one decay per second, or 1 s-1
  • This equation shows:
    • The greater the decay constant, the greater the activity of the sample
    • The activity depends on the number of undecayed nuclei remaining in the sample
    • The minus sign indicates that the number of nuclei remaining decreases with time
  • Half-life and decay constant can be linked, using an equation called the exponential decay equation

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

  • This equation shows how the number of undecayed nuclei, N, changes over time, t, where N0 is the initial number of nuclei in the sample
  • When time t is equal to the half-life t½, the number of undecayed nuclei in the sample, N, will fall to half of its original value open parentheses N space equals space 1 half N subscript 0 close parentheses

1 half N subscript 0 space equals space N subscript 0 space e to the power of negative lambda t subscript 1 divided by 2 end subscript end exponent

  • The formula linking half-life and decay constant can then be derived as follows:

divide both sides by N01 half space equals space e to the power of negative lambda t subscript 1 divided by 2 end subscript end exponent

take the natural log of both sides:  ln space open parentheses 1 half close parentheses space equals space minus lambda t subscript 1 divided by 2 end subscript

apply properties of logarithms:  lambda t subscript 1 divided by 2 end subscript space equals space ln space 2

  • Therefore, half-life t½ can be calculated using the equation:

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

  • This equation shows that half-life t½ and the radioactive decay rate constant λ are inversely proportional
    • Therefore, the shorter the half-life, the larger the decay constant and the faster the decay
  • The half-life of a radioactive substance can be determined from decay curves and log graphs
  • Since half-life is the time taken for the initial number of nuclei, or activity, to reduce by half, it can be found by
    • Drawing a line to the curve at the point where the activity has dropped to half of its original value
    • Drawing a line from the curve to the time axis, this is the half-life

Half Life Decay Curves 1, downloadable AS & A Level Physics revision notes

A linear decay curve. This represents the relationship: N space equals space N subscript 0 space e to the power of negative lambda t end exponent

  • Straight-line graphs tend to be more useful than curves for interpreting data
    • Due to the exponential nature of radioactive decay, logarithms can be used to achieve a straight-line graph
  • Take the exponential decay equation for the number of nuclei

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

  • Taking the natural logs of both sides

ln space N space equals space ln space N subscript 0 space minus space lambda t

ln space N space equals negative lambda t space plus space ln space N subscript 0 space

  • In this form, this equation can be compared to the equation of a straight line

y space equals space m x space plus space c

  • Where:
    • ln N is plotted on the y-axis
    • t is plotted on the x-axis
    • gradient = −λ
    • y-intercept = ln N0
  • Half-lives can be found in a similar way to the decay curve but the intervals will be regular as shown below:

Half Life Decay Curves 2, downloadable AS & A Level Physics revision notes

A logarithmic graph. This represents the relationship: ln space N equals negative lambda t plus ln space N subscript 0

  • Note: experimentally, the measurement generally taken is the count rate of the source
    • Since count rate ∝ activity ∝ number of nuclei, the graphs will all take the same shapes when plotted against time (or number of half-lives) linearly or logarithmically  

Worked example

Radium is a radioactive element first discovered by Marie and Pierre Curie.

They used the radiation emitted from radium-226 to define a unit called the Curie (Ci) which they defined as the activity of 1 gram of radium.

It was found that in a 1 g sample of radium, 2.22 × 1012 atoms decayed in 1 minute.

Another sample containing 3.2 × 1022 radium-226 atoms had an activity of 12 Ci.

(a) Determine the value of 1 Curie

(b) Determine the decay constant for radium-226

Answer:

(a)

Step 1: Write down the known quantities

  • Number of atoms decayed, ΔN = 2.22 × 1012
  • Time, Δt = 1 minutes = 60 s

Step 2: Write down the activity equation

Step 3: Calculate the value of 1 Ci

(b)

Step 1: Write down the known quantities

  • Number of atoms, N = 3.2 × 1022
  • Activity, A = 12 Ci = 12 × (3.7 × 1010) = 4.44 × 1011 Bq

Step 2: Write down the activity equation

A = λN

Step 3: Calculate the decay constant of radium

  • Therefore, the decay constant of radium-226 is 1.4 × 10–11 s–1

Worked example

Strontium-90 is a radioactive isotope with a half-life of 28.0 years. A sample of Strontium-90 has an activity of 6.4 × 109 Bq.

(a)
Calculate the decay constant λ, in year–1, of Strontium-90.
(b)
Determine the fraction of the sample remaining after 50 years.

 

Answer:

(a)

Step 1: List the known quantities

  • Half-life, t½ = 28 years

Step 2: Write the equation for half-life

t subscript bevelled 1 half end subscript equals fraction numerator ln space 2 over denominator lambda end fraction

Step 3: Rearrange for λ and calculate

lambda equals fraction numerator ln space 2 over denominator t subscript bevelled 1 half end subscript end fraction equals fraction numerator ln space 2 over denominator 28 end fraction= 0.025 year−1

(b)

Step 1: List the known quantities

  • Decay constant, λ = 0.025 year−1
  • Time passed, t = 50 years

Step 2: Write the equation for exponential decay

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

Step 3: Rearrange for N over N subscript 0 and calculate

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

N over N subscript 0 equals e to the power of negative left parenthesis 0.025 right parenthesis cross times 50 end exponent = 0.287

  • Therefore, 28.7% of the sample will remain after 50 years

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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.