Radioactive decay law N = N₀e^(-λt) — derive and solve numerical

Question

Derive the radioactive decay law $N = N_0 e^{-\lambda t}$ . A radioactive sample has a half-life of 20 minutes. Find (a) the decay constant $\lambda$ , and (b) the fraction of the sample remaining after 1 hour.

(JEE Main 2023, similar pattern)

Solution — Step by Step

The rate of decay is proportional to the number of undecayed nuclei at any time:

\frac{dN}{dt} = -\lambda N

The negative sign indicates that $N$ decreases with time. $\lambda$ is the decay constant (probability of decay per unit time per nucleus).

\frac{dN}{N} = -\lambda \, dt

Integrating both sides:

\ln N = -\lambda t + C

At $t = 0$ , $N = N_0$ , so $C = \ln N_0$ .

\ln\frac{N}{N_0} = -\lambda t

\boxed{N = N_0 e^{-\lambda t}}

At $t = T_{1/2}$ (half-life), $N = N_0/2$ :

\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{20} = \mathbf{0.03465 \text{ min}^{-1}}

1 hour = 60 minutes = 3 half-lives (since $T_{1/2} = 20$ min).

After each half-life, the sample halves:

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

So $\frac{1}{8}$ or 12.5% of the original sample remains.

Why This Works

Radioactive decay is a random process at the individual nucleus level, but statistically predictable for large numbers. The exponential law emerges because each nucleus has the same probability $\lambda \, dt$ of decaying in a small time interval, independent of how long it has already existed. This “memoryless” property naturally leads to the exponential function.

The half-life is a more intuitive measure than $\lambda$ — it tells us the time for the activity to drop by half. After $n$ half-lives, the fraction remaining is $(1/2)^n$ .

Alternative Method — Using Activity

Activity $A = \lambda N = A_0 e^{-\lambda t}$ , where $A_0 = \lambda N_0$ .

For the numerical: $A$ after 60 min = $A_0 \times e^{-0.03465 \times 60} = A_0 \times e^{-2.079} = A_0/8$ .

When the time is an exact multiple of half-life, skip the exponential calculation entirely. Just halve repeatedly: after 1 half-life → $1/2$ , after 2 → $1/4$ , after 3 → $1/8$ , after 4 → $1/16$ . This saves precious time in JEE/NEET.

Common Mistake

Students often write $N = N_0 / 3$ after 3 half-lives instead of $N = N_0 / 8$ . The decay is exponential, not linear. Each half-life halves the remaining sample, not the original. After 3 half-lives: $N_0 \to N_0/2 \to N_0/4 \to N_0/8$ . Not $N_0 \to 2N_0/3 \to N_0/3 \to 0$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Activity

Common Mistake

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