MCAT content library / Chemistry & physics / Radioactive decay and semilog data
Read half-life from count data, distinguish fractional from absolute loss, and recognize why exponential decay becomes linear on an appropriate semilog plot.
About 15 minutes, including the background-corrected count example and checks.
Useful foundations
On this page
- Radioactive decay removes a constant fraction, not a constant amount
- Concept: half-life and decay constant describe the same timescale
- Worked application: read half-life from a count table
- Pattern: a semilog line reveals exponential decay
- Rule: background must be subtracted before fitting the decay
- Check your reasoning
- Continue learning
- Sources
Radioactive decay removes a constant fraction, not a constant amount
For a sample of unstable nuclei with a constant decay probability per unit time, the expected number remaining follows exponential decay:
N(t) = N₀e−λt.
The same relationship can be written using half-life:
N(t) = N₀(1/2)t/t₁⁄₂.
After each half-life, half of the nuclei present at the start of that interval remain. The fraction lost is constant, but the absolute number lost becomes smaller as the sample shrinks.
Concept: half-life and decay constant describe the same timescale
The decay constant λ and half-life are related by
t₁⁄₂ = ln 2 / λ.
A larger λ means a greater probability of decay per unit time and therefore a shorter half-life. Half-life does not depend on the initial amount of a particular isotope under the ordinary decay model.
Two samples of the same isotope can begin with very different activities or numbers of nuclei but still lose the same fraction over one half-life.
Worked application: read half-life from a count table
An original detector experiment measures a radioactive sample after background has been subtracted:
| Time (h) | Background-corrected count rate |
|---|---|
| 0 | 800 counts/min |
| 3 | 400 counts/min |
| 6 | 200 counts/min |
| 9 | 100 counts/min |
| 12 | 50 counts/min |
The count rate halves every 3 hours, so the measured half-life is 3 h. Between 0 and 3 h, the rate falls by 400 counts/min. Between 3 and 6 h, it falls by only 200 counts/min. The absolute losses differ, but each interval removes 50% of what was present at its start.
After 12 h, four half-lives have elapsed, so the remaining fraction is (1/2)4 = 1/16, matching 50/800.
Pattern: a semilog line reveals exponential decay
Taking the natural logarithm gives:
ln N = ln N₀ − λt.
A graph of ln N versus time is therefore a straight line with slope −λ for ideal single-exponential decay. A semilog plot can make an exponential process easier to recognize because equal fractional changes become equal vertical intervals.
A straight semilog line does not mean the sample loses a constant absolute number of nuclei each hour. It means the logarithm changes linearly while the original count changes exponentially.
Rule: background must be subtracted before fitting the decay
A detector can record environmental or instrument background even when the sample contribution is small. If the measured rate is sample + background, the raw curve may appear to flatten toward the background level at late times.
Suppose the actual sample signal is 50 counts/min but background is 30 counts/min. The detector reads 80 counts/min. Treating 80 as pure sample activity would overestimate the amount remaining and distort the fitted half-life. Measure or otherwise estimate the background and subtract it before applying the simple decay model.
ACTIVE PRACTICE · PHYSICAL SCIENCE
Predict the Fraction Remaining
The lesson’s background-corrected count rate starts at 800 counts/min and halves every 3 hours. Predict the measurement after 9 hours before manipulating the exponential model.
Predicted background-corrected count rate: counts/min; remaining fraction; elapsed half-lives.
The graph uses expected decay behavior; measured radioactive counts fluctuate. Instrument background is assumed already subtracted.
Check your reasoning
Choose an answer before opening its feedback.
Check 1
A sample has a half-life of 4 h. What fraction remains after 12 h?
A. 1/8
Correct. Twelve hours is three half-lives: (1/2)³ = 1/8.
B. 1/3
Review this choice. Half-life produces repeated halving, not linear subtraction by thirds.
C. 1/4
Review this choice. One quarter remains after two half-lives, not three.
D. 0
Review this choice. Exponential decay approaches zero but does not reach it after a finite number of half-lives in the ideal model.
Check 2
A sample falls from 1000 nuclei to 500 in one half-life. During the next half-life, how many nuclei are expected to decay?
A. 250
Correct. Half of the 500 nuclei present at the start of the second interval decay.
B. 500
Review this choice. That was the first interval’s absolute loss.
C. 750
Review this choice. The sample contains only 500 nuclei at the start of the second interval.
D. 1000
Review this choice. The original sample size is no longer present.
Check 3
An ideal single-isotope decay gives a straight line on a plot of ln(count) versus time. What does the slope represent?
A. −λ
Correct. ln N = ln N₀ − λt.
B. The initial count N₀
Review this choice. ln N₀ is the intercept.
C. The detector background only
Review this choice. Background should be corrected before fitting the simple model.
D. The atomic mass number
Review this choice. Mass number is not the semilog decay slope.