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Exponential Decay (AQA A Level Physics)

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Katie M

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Katie M

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Exponential 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 = N0 e–λt

  • 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 = A0 e–λt

  • 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 = C0 e–λt

  • Where:

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

    • C0 = initial count rate (counts per minute or cpm)

The exponential function e

  • The symbol e represents the exponential constant

    • It is approximately equal to e = 2.718

  • On a calculator it is shown by the button ex

  • The inverse function of ex is ln(y), known as the natural logarithmic function

    • This is because, if ex = y, then x = ln(y)

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 = A0 e–λt

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

The Exponential Nature of Radioactive Decay Worked Example equation 1

Step 4: Calculate the ratio A/A0

The Exponential Nature of Radioactive Decay Worked Example equation 2

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

Using Molar Mass & The Avogadro Constant

Molar Mass

  • The molar mass, or molecular mass, of a substance is the mass of a substance, in grams, in one mole

    • Its unit is g mol-1

  • The number of moles from this can be calculated using the equation:

The Avogadro Constant equation 2

Avogadro’s Constant

  • Avogadro’s constant (NA) is defined as:

    The number of atoms in one mole of a substance; equal to 6.02 × 1023 mol-1

  • For example, 1 mole of sodium (Na) contains 6.02 × 1023 atoms of sodium

  • The number of atoms, N, can be determined using the equation:

Activity & The Decay Constant Worked Example equation 1

Worked Example

Americium-241 is an artificially produced radioactive element that emits α-particles.

In a smoke detector, a sample of americium-241 of mass 5.1 µg is found to have an activity of 5.9 × 105 Bq. The supplier’s website says the americium-241 in their smoke detectors initially has an activity level of 6.1 × 105 Bq.

Determine:

(a) the number of nuclei in the sample of americium-241

(b) the decay constant of americium-241

(c) the age of the smoke detector in years.

Answer:

Part (a)

Step 1: Write down the known quantities

  • Mass = 5.1 μg = 5.1 × 10−6 g

    • Molecular mass of americium = 241

    • Avogadro constant, N subscript A = 6.02 × 1023 mol−1

Step 2: Write down the equation relating number of nuclei, mass and molecular mass

n u m b e r space o f space n u c l e i comma space 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

Step 3: Calculate the number of nuclei

number of nuclei, N space equals space fraction numerator open parentheses 5.1 cross times 10 to the power of negative 6 end exponent close parentheses cross times open parentheses 6.02 cross times 10 to the power of 23 close parentheses over denominator 241 end fraction = 1.27 × 1016

Part (b)

Step 1: Write down the known quantities

  • Activity, A = 5.9 × 105 Bq

  • Number of nuclei, N = 1.27 × 1016

Step 2: Write the equation for activity

Activity, A space equals space lambda N

Step 3: Rearrange for decay constant λ and calculate the answer

lambda space equals space fraction numerator space A over denominator N end fraction space equals space fraction numerator 5.9 cross times 10 to the power of 5 over denominator 1.27 cross times 10 to the power of 16 end fraction space equals space 4.65 cross times 10 to the power of negative 11 end exponent space s to the power of negative 1 end exponent

Part (c)

Step 1: Write down the known quantities

  • Activity, A = 5.9 × 105 Bq

  • Initial activity, A0 = 6.1 × 105 Bq

  • Decay constant, λ = 4.65 × 10–11 s–1

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 for time t

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

ln space open parentheses A over A subscript 0 close parentheses space equals space minus lambda t

t space equals space minus 1 over lambda ln space open parentheses A over A subscript 0 close parentheses space

Step 4: Calculate the age of the smoke detector and convert to years

t space equals space minus fraction numerator 1 over denominator 4.65 cross times 10 to the power of negative 11 end exponent end fraction space ln space open parentheses fraction numerator 5.9 cross times 10 to the power of 5 over denominator 6.1 cross times 10 to the power of 5 end fraction close parentheses space equals space 7.169 cross times 10 to the power of 8 s

t space equals space fraction numerator 7.169 cross times 10 to the power of 8 over denominator 24 cross times 60 cross times 60 cross times 365 end fraction = 22.7 years

  • Therefore, the smoke detector is 22.7 years old

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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.