N = N₀(1/2)^(t/t½). After 3 half-lives, what fraction remains?

Radioactive Decay — Half-Life Problems

Question

A radioactive sample has a half-life of 20 years. After 60 years, what fraction of the original sample remains?

Solution — Step by Step

Half-life $t_{1/2} = 20$ years. Total time elapsed $t = 60$ years. We need the fraction $\frac{N}{N_0}$ remaining.

The key move here is converting total time into number of half-lives — the formula becomes trivially easy after that.

n = \frac{t}{t_{1/2}} = \frac{60}{20} = 3 \text{ half-lives}

After each half-life, the sample halves. After $n$ half-lives:

\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8}

After 60 years (3 half-lives), one-eighth (1/8) of the original sample remains.

If you started with $N_0 = 800$ atoms, you’d have $800 \times \frac{1}{8} = 100$ atoms left.

Why This Works

Every radioactive nucleus has a fixed probability of decaying per unit time — that’s the defining property of radioactive decay. Because decay is statistical, the fraction surviving in any fixed time interval is always the same, regardless of how much material you started with.

That’s why we get an exponential: after one half-life, half survives. After another, half of that survives. We’re repeatedly multiplying by $\frac{1}{2}$ , which is exactly $\left(\frac{1}{2}\right)^n$ .

This is why the half-life formula works even when $n$ isn’t a whole number — it’s just the same exponential evaluated at a fractional power.

Alternative Method

Use the full exponential decay law: $N = N_0 e^{-\lambda t}$ , where $\lambda = \frac{0.693}{t_{1/2}}$ .

\lambda = \frac{0.693}{20} = 0.03465 \text{ yr}^{-1}

N = N_0 \, e^{-0.03465 \times 60} = N_0 \, e^{-2.079}

Now, $e^{-2.079} = e^{-\ln 8} = \frac{1}{8}$ — same answer.

For CBSE and JEE Main, the $\left(\frac{1}{2}\right)^n$ method is faster whenever $t$ is a clean multiple of $t_{1/2}$ . Save the $e^{-\lambda t}$ form for when it isn’t — like “after 30 years” in this problem.

Common Mistake

Many students write the fraction remaining as $\frac{3}{4}$ or $\frac{1}{4}$ — they confuse “3 half-lives” with “3/4 decayed, 1/4 left.” Track it step by step: after 1st half-life → $\frac{1}{2}$ remains; after 2nd → $\frac{1}{4}$ ; after 3rd → $\frac{1}{8}$ . The fraction left is always $\left(\frac{1}{2}\right)^n$ , not $\frac{n}{2^n}$ or any other misread formula.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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