Half-Life Formula: Worked Examples for Decay Problems

A sample's activity reads 1200 becquerels at noon. The isotope has a half-life of 6.0 hours. What will the detector read at 6 a.m. the next day? The answer is not 1200 minus three equal steps. Each half-life removes half of what is left, so the readings run 1200, 600, 300, 150. That multiplicative pattern is the whole content of the half-life formula.

This guide works the formula from the simple whole-number case to the cases that trip people up on exams: a fractional number of half-lives, finding the half-life from two measurements, switching to the decay constant, computing activity from a number of nuclei, and dating a sample from its remaining carbon-14. The Nuclear Physics topic lists the decay skills these examples practise.

Two forms of the same decay law

Radioactive decay is random for one nucleus but very regular for a large number of them. If $N_0$ undecayed nuclei are present at the start and the half-life is $T_{1/2}$, the number left after time $t$ is

$$N(t)=N_0\left(\frac{1}{2}\right)^{t/T_{1/2}}.$$

The exponent $t/T_{1/2}$ is simply the number of half-lives that have passed. It does not have to be a whole number.

Textbooks also write the law with the decay constant $\lambda$. OpenStax University Physics, Volume 3, §10.3 gives

$$N(t)=N_0e^{-\lambda t},\qquad \lambda=\frac{\ln 2}{T_{1/2}}\approx\frac{0.693}{T_{1/2}}.$$

The two forms agree because $\left(\frac{1}{2}\right)^{t/T_{1/2}}=e^{-(\ln 2)t/T_{1/2}}$. Use the half-life form when the question talks in half-lives, and the exponential form when it gives or asks for $\lambda$. The same relation holds for the mass of the parent isotope and for its activity, because each is proportional to $N$.

Before any arithmetic, put $t$ and $T_{1/2}$ in the same unit. Hours divided by days gives a number that looks reasonable and is wrong.

Example 1: a whole number of half-lives

Return to the opening sample: $A_0=1200\ \mathrm{Bq}$ and $T_{1/2}=6.0\ \mathrm h$. From noon to 6 a.m. is 18 h, which is $18/6.0=3$ half-lives.

$$A=1200\ \mathrm{Bq}\times\left(\frac{1}{2}\right)^{3}=\frac{1200\ \mathrm{Bq}}{8}=150\ \mathrm{Bq}.$$

A quick table makes the pattern visible and catches arithmetic slips:

The tempting wrong answer subtracts 600 Bq every 6 h and reaches zero after 12 h. Decay is proportional to what remains, so the steps shrink.

Example 2: a fractional number of half-lives

A hypothetical isotope has a half-life of 8.0 days, and a source starts with 40 mg of it. How much of that isotope remains after 20 days?

The number of half-lives is $20/8.0=2.5$. The formula handles that directly:

$$m=40\ \mathrm{mg}\times\left(\frac{1}{2}\right)^{2.5}\approx40\ \mathrm{mg}\times0.177\approx7.1\ \mathrm{mg}.$$

Check the size: after 2 half-lives there would be 10 mg, after 3 half-lives 5 mg. A result between 5 and 10 mg is right, and it is closer to 5 than to 10 because the curve falls fastest at the start of each interval. Halfway between 10 and 5, 7.5 mg, would be a linear guess, not the decay law.

The 7.1 mg is the remaining parent isotope. The sample does not lose 33 mg of total mass; most of the decayed atoms are still there as daughter nuclei.

Example 3: find the half-life from two readings

A detector shows 800 Bq for a pure source, and 3.0 h later it shows 500 Bq after background has been subtracted. What is the half-life?

Solve the half-life form for the number of half-lives:

$$\frac{t}{T_{1/2}}=\log_2\left(\frac{A_0}{A}\right)=\log_2(1.6)=\frac{\ln 1.6}{\ln 2}\approx0.678.$$

Then

$$T_{1/2}=\frac{3.0\ \mathrm h}{0.678}\approx4.4\ \mathrm h.$$

Sanity check: the activity fell by less than half in 3.0 h, so the half-life must be longer than 3.0 h. If you get a negative half-life, or one shorter than the elapsed time while the reading has not yet halved, the ratio was inverted.

When the drop is a clean power of two, skip the logarithm. A fall from 960 Bq to 240 Bq in 10 h is a factor of 4, which is 2 half-lives, so $T_{1/2}=5.0\ \mathrm h$.

Example 4: decay constant and activity from a number of nuclei

A source contains $N=2.0\times10^{15}$ nuclei of an isotope with $T_{1/2}=8.0$ days. What is its activity in becquerels?

Activity is decays per second, so convert the half-life to seconds first: $8.0\ \mathrm{d}=8.0\times86400\ \mathrm s=6.91\times10^{5}\ \mathrm s$. Then

$$\lambda=\frac{\ln 2}{T_{1/2}}=\frac{0.693}{6.91\times10^{5}\ \mathrm s}\approx1.0\times10^{-6}\ \mathrm{s^{-1}},$$ $$A=\lambda N\approx1.0\times10^{-6}\ \mathrm{s^{-1}}\times2.0\times10^{15}\approx2.0\times10^{9}\ \mathrm{Bq}.$$

Had you left the half-life in days, the result would come out in decays per day, about 86 400 times larger than the number of becquerels. The unit of $\lambda$ decides the unit of $A$.

Example 5: carbon dating with a non-round answer

OpenStax gives carbon-14 a half-life of 5730 years. A wooden object has 30% of the carbon-14 activity per gram of comparable living wood. Assuming that starting ratio, how old is it?

$$t=T_{1/2}\log_2\left(\frac{A_0}{A}\right)=5730\times\log_2\left(\frac{1}{0.30}\right)\approx5730\times1.74\approx9.95\times10^{3}.$$

The object is roughly 10 000 years old. Check the bracket: 25% remaining would mean exactly 2 half-lives, 11 460 years. Thirty percent is a little more than 25%, so the age must be a little less, and it is.

The answer depends on the assumption that the starting ratio matched living material. Real radiocarbon laboratories calibrate for changes in atmospheric carbon-14, so treat this as a physics exercise, not a laboratory date.

Mistakes that cost marks

• Subtracting equal amounts. Each half-life halves the current amount. Linear steps reach zero; the decay law never does.
• Mixed time units. $t$ and $T_{1/2}$ must match, and $\lambda$ must be in $\mathrm{s^{-1}}$ if you want activity in becquerels.
• Mixing up log and ln. $\ln 2\approx0.693$, while $\log_{10}2\approx0.301$. If you take $\log_{10}$ of the ratio, divide by $\log_{10}2$, not by 0.693. Our article on physics mistakes that are really math mistakes has more of these.
• Confusing parent and sample. The formula tracks the undecayed parent nuclei, not the total mass of the sample.
• Ignoring background. Count rates measured in a lab include background radiation. Subtract it before taking ratios.

Three practice problems with answers

Problem 1. An isotope with $T_{1/2}=15\ \mathrm h$ starts at 6400 Bq. What is the activity after 60 h? Four half-lives: $6400/16=400\ \mathrm{Bq}$.

Problem 2. What fraction of a sample remains after 1.5 half-lives? $\left(\frac{1}{2}\right)^{1.5}\approx0.354$, about 35%, not 25%.

Problem 3. A count rate falls from 1000 to 125 counts per minute, background already removed, in 36 min. What is the half-life? The ratio 8 is 3 half-lives, so $T_{1/2}=12\ \mathrm{min}$.

Half-life questions appear in almost every school physics course; our A-level Physics guide shows where radioactive decay sits among the other modern physics skills. Whatever the course, the routine is the same: match the units, count half-lives as $t/T_{1/2}$, let the exponent be fractional when it needs to be, and check that the answer falls between the whole-half-life values around it.