Half-Life, Decay Law, Activity, and Units Used in Radioactivity

Why radioactive decay follows an exponential law

Radioactive decay is random for each nucleus but predictable for a large collection. The key idea is that each undecayed nucleus has the same constant probability per unit time of decaying. This probability does not depend on how long that nucleus has already existed: the process is memoryless.

Conceptual derivation (from constant probability to an exponential)

Let N(t) be the number of undecayed nuclei at time t. Assume that in a short time interval Δt, a fixed fraction of the remaining nuclei decays. That means the expected change is proportional to how many are still present:

ΔN ≈ −(constant) · N · Δt

Define the proportionality constant as the decay constant λ (lambda), with units of 1/time. Then:

ΔN ≈ −λ N Δt

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Divide by Δt and take the limit as Δt → 0 to obtain the differential equation:

dN/dt = −λN

This equation says: the rate of decrease is proportional to how many remain. Solving it gives the exponential decay law:

N(t) = N0 e^{−λt}

where N0 is the initial number of nuclei at t = 0.

Half-life as a statistical property

The half-life T1/2 is the time required for the expected number of undecayed nuclei to drop to half its initial value:

N(T1/2) = N0/2

Using N(t) = N0 e^{−λt}:

N0/2 = N0 e^{−λT1/2} → 1/2 = e^{−λT1/2}

Take natural logs:

ln(1/2) = −λT1/2

Since ln(1/2) = −ln 2:

T1/2 = (ln 2)/λ and equivalently λ = (ln 2)/T1/2

Statistical meaning: half-life does not mean each nucleus “waits” exactly T1/2 and then decays. Individual decay times vary widely; half-life describes the behavior of a large ensemble (and the expected value for the fraction remaining).

Computing remaining quantity and fraction remaining

Using half-life directly (repeated halving)

After n half-lives, the remaining fraction is:

f = (1/2)^n

and the remaining number is:

N = N0 (1/2)^n

where n = t / T1/2 (not necessarily an integer).

Using logarithms (for non-integer numbers of half-lives)

Combine N(t) = N0 e^{−λt} with λ = ln2/T1/2:

N(t) = N0 e^{−(ln2)t/T1/2} = N0 · 2^{−t/T1/2}

So the fraction remaining is:

f = N/N0 = 2^{−t/T1/2}

To solve for time given a remaining fraction f:

f = 2^{−t/T1/2} → log2(f) = −t/T1/2 → t = −T1/2 · log2(f)

Using natural logs (often more convenient):

t = −(T1/2/ln2) · ln(f)

Worked examples: remaining amount, fraction remaining, elapsed time

Example 1: integer number of half-lives

A sample starts with N0 = 8.0×10^20 nuclei. The half-life is T1/2 = 5.0 days. How many remain after t = 20 days?

• Step 1: Compute number of half-lives: n = t/T1/2 = 20/5 = 4.
• Step 2: Remaining fraction: f = (1/2)^4 = 1/16.
• Step 3: Remaining nuclei: N = N0/16 = (8.0×10^20)/16 = 5.0×10^19.

Example 2: non-integer half-lives (use exponent form)

A radionuclide has T1/2 = 6.0 h. What fraction remains after t = 10.0 h?

• Step 1: Compute t/T1/2 = 10/6 = 1.6667.
• Step 2: Use f = 2^{−t/T1/2}: f = 2^{−1.6667}.
• Step 3: Evaluate: 2^{−1.6667} = 1/(2^{1.6667}) ≈ 1/3.17 ≈ 0.315.

About 31.5% of the nuclei remain.

Example 3: find elapsed time from remaining fraction (logarithms)

A sample has half-life T1/2 = 12.0 years. How long until only 10% remains?

• Step 1: Set f = 0.10.
• Step 2: Use t = −(T1/2/ln2) ln(f).
• Step 3: Compute: t = −(12.0/0.693) ln(0.10).
• Step 4: Since ln(0.10) = −2.3026, t = (12.0/0.693)·2.3026 ≈ 17.32·2.3026 ≈ 39.9 years.

It takes about 40 years to reach 10% remaining.

Example 4: time to drop from one amount to another (ratio method)

A sample decreases from N1 to N2. Show that time depends only on the ratio N2/N1.

From N(t) = N0 e^{−λt}, for two times:

N2/N1 = e^{−λ(t2−t1)}

So the elapsed time Δt = t2−t1 is:

Δt = (1/λ) ln(N1/N2)

Example: If N drops by a factor of 5, then Δt = (1/λ) ln 5, regardless of the starting amount.

Activity, decay constant, and what units mean physically

Definition of activity

The activity A is the number of decays per unit time. It is proportional to how many nuclei are available to decay:

A = λN

This follows directly from dN/dt = −λN: the decay rate magnitude is −dN/dt = λN, which is the activity.

How activity changes with time

Since N(t) decays exponentially, activity also decays exponentially:

A(t) = λN0 e^{−λt} = A0 e^{−λt}

So activity has the same half-life as the number of nuclei: after one half-life, the activity halves.

Units: Bq and Ci

• Becquerel (Bq): 1 Bq = 1 decay per second (physically: one nuclear transformation each second).
• Curie (Ci): 1 Ci = 3.7×10^10 decays per second = 3.7×10^10 Bq.

Important interpretation: activity measures how many decays occur per second, not the energy per decay, not the dose, and not the “danger” by itself. Two sources can have the same activity but different radiation types/energies.

Worked examples: activity calculations and unit conversions

Example 5: compute activity from N and half-life

A sample contains N = 2.0×10^15 radioactive nuclei with half-life T1/2 = 30.0 days. Find the activity in Bq.

• Step 1: Convert half-life to seconds: 30.0 days = 30.0×24×3600 = 2.592×10^6 s.
• Step 2: Compute decay constant: λ = ln2/T1/2 = 0.693/(2.592×10^6) = 2.67×10^−7 s^−1.
• Step 3: Activity: A = λN = (2.67×10^−7)(2.0×10^15) = 5.34×10^8 s^−1.
• Step 4: Convert to Bq: 5.34×10^8 s^−1 = 5.34×10^8 Bq.

Example 6: convert Bq to Ci

Convert 7.4×10^10 Bq to curies.

• Use 1 Ci = 3.7×10^10 Bq.
• Activity in Ci = (7.4×10^10)/(3.7×10^10) = 2.0 Ci.

Example 7: compare activities of two samples of the same isotope

Two samples contain the same radionuclide (same λ). Sample A has N_A = 3.0×10^12 nuclei; sample B has N_B = 7.5×10^12 nuclei. Compare activities.

Because A = λN and λ is the same:

A_B/A_A = N_B/N_A = (7.5×10^12)/(3.0×10^12) = 2.5

Sample B has 2.5× the activity of sample A.

Example 8: compare activities of equal numbers of nuclei with different half-lives

Two samples each contain N = 1.0×10^14 nuclei. Isotope X has T1/2 = 1.0 hour, isotope Y has T1/2 = 10.0 hours. Which has higher activity and by what factor?

Compute ratio using A = λN and λ = ln2/T1/2:

A_X/A_Y = λ_X/λ_Y = (ln2/1 h)/(ln2/10 h) = 10

Isotope X has 10× higher activity because it decays faster.

Mixed practice: half-life ↔ decay constant, activity comparisons, and graphs

Practice set A: convert between half-life and decay constant

• A1. A nuclide has T1/2 = 2.5 min. Find λ in s^−1.
  T1/2 = 150 s, so λ = 0.693/150 = 4.62×10^−3 s^−1.
• A2. A nuclide has λ = 1.20×10^−6 s^−1. Find T1/2 in days.
  T1/2 = 0.693/λ = 0.693/(1.20×10^−6) = 5.78×10^5 s.
  Convert: 5.78×10^5 s /(86400 s/day) = 6.69 days.

Practice set B: compute activity and remaining amount together

A source has initial activity A0 = 1.6×10^6 Bq and half-life T1/2 = 4.0 h.

• B1. Activity after t = 12 h: t/T1/2 = 3, so A = A0(1/2)^3 = (1.6×10^6)/8 = 2.0×10^5 Bq.
• B2. Fraction of nuclei remaining after 12 h is the same as activity fraction: f = 1/8 = 0.125.

Interpreting activity–time graphs

On an activity–time plot, exponential decay has a distinctive shape:

• Linear axes (A vs t): a curve that drops quickly at first and then flattens. Equal time steps do not correspond to equal drops in activity.
• Semi-log plot (ln A vs t): a straight line because ln A = ln A0 − λt. The slope is −λ and the intercept is ln A0.

Practice set C: reading half-life from an activity-time description

A graph shows activity dropping from 800 Bq to 200 Bq over 6 hours. Determine the half-life.

• The activity decreased by a factor of 800/200 = 4.
• A factor of 4 corresponds to two half-lives because (1/2)^2 = 1/4.
• So 6 hours equals 2 half-lives, giving T1/2 = 3 hours.

Practice set D: solve for time using activity ratio

A source’s activity drops from A1 to A2. Since A(t) = A0 e^{−λt}, the elapsed time is:

Δt = (1/λ) ln(A1/A2)

Example: A nuclide has T1/2 = 8.0 days. How long for activity to drop from 5.0 MBq to 1.0 MBq?

• Compute λ = ln2/T1/2 = 0.693/8.0 = 0.0866 day^−1.
• Use ratio method: Δt = (1/0.0866) ln(5.0/1.0) = 11.55·ln5.
• ln5 = 1.609, so Δt ≈ 11.55×1.609 = 18.6 days.