Activity of sample drops to 1/8 in 30 days — find half-life

Question

The activity (radioactivity) of a sample drops to 1/8 of its initial value in 30 days. Find the half-life of the radioactive substance.

Solution — Step by Step

Activity ( $A$ ) of a radioactive sample is directly proportional to the number of undecayed nuclei ( $N$ ). So if activity drops to $A_0/8$ , the number of nuclei has also dropped to $N_0/8$ . The fraction remaining is $1/8$ .

After $n$ half-lives, the fraction remaining is $(1/2)^n$ .

We need: $(1/2)^n = 1/8 = (1/2)^3$

Therefore $n = 3$ — exactly 3 half-lives have elapsed.

Total time = 30 days; number of half-lives = 3.

T_{1/2} = \frac{\text{Total time}}{n} = \frac{30}{3} = \mathbf{10\text{ days}}

The half-life of the substance is 10 days.

Why This Works

Activity $A = \lambda N$ where $\lambda$ is the decay constant. Since both $A$ and $N$ decay with the same time constant, a reduction in activity by a factor of $f$ is identical to a reduction in the number of nuclei by the same factor. This means the half-life (the time for half the nuclei to decay) is also the time for activity to halve.

So “activity drops to 1/8” = “sample reduces to 1/8” = “3 half-lives have elapsed.”

Alternative Method

Using the exponential decay law: $A = A_0 e^{-\lambda t}$ .

$A_0/8 = A_0 e^{-\lambda \times 30}$

$e^{-\lambda \times 30} = 1/8 = e^{-\ln 8}$

$\lambda \times 30 = \ln 8 = 3\ln 2$

$\lambda = \frac{3\ln 2}{30} = \frac{\ln 2}{10}$

$T_{1/2} = \frac{\ln 2}{\lambda} = 10$ days ✓

The key recognition step: $1/8 = (1/2)^3$ , so 3 half-lives. Always write $1/f$ as $(1/2)^n$ and read off $n$ . Common fractions and their half-lives: $1/2 \to 1$ ; $1/4 \to 2$ ; $1/8 \to 3$ ; $1/16 \to 4$ ; $1/32 \to 5$ . These are the cleanest exam numbers.

Common Mistake

Students sometimes confuse “activity drops to 1/8” with “7/8 of the activity has been lost.” These are the same thing mathematically, but students sometimes use the wrong fraction: they compute half-life = 30/(7/8) which is nonsensical. Always use the fraction remaining (1/8), not the fraction lost (7/8), in the half-life calculation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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