What Radioactive Decay Means
Radioactivity, discovered by A. H. Becquerel in 1896, was later shown to be a nuclear phenomenon in which an unstable nucleus spontaneously transforms by emitting radiation. NCERT identifies three modes of natural radioactive decay: $\alpha$-decay, in which a helium nucleus is emitted; $\beta$-decay, in which electrons or positrons are emitted; and $\gamma$-decay, in which high-energy photons are released. In each case the parent nucleus moves toward a more stable configuration.
The defining feature of this process is that it is spontaneous and random at the level of a single nucleus. One cannot predict when any particular nucleus will decay. The NIOS text emphasises that the rate of disintegration is independent of external factors such as temperature, pressure or chemical state, and depends only on the laws of chance. What makes the phenomenon quantitatively tractable is that a real sample contains an enormous number of nuclei, and the statistics of large numbers produce a perfectly smooth, predictable decay curve.
This combination of microscopic randomness and macroscopic regularity is exactly what the law of radioactive decay captures. The unpredictability of the individual is washed out into the certainty of the average, and that average follows a single exponential function of time.
Decay rate cannot be sped up or slowed down
A common misconception is that heating, cooling, applying pressure, or changing the chemical compound can alter how fast a sample decays. It cannot. The decay constant $\lambda$ is an intrinsic property of the nuclide alone.
Remember: $\lambda$, and therefore $T_{1/2}$ and $\tau$, are fixed for a given radioactive species, independent of physical or chemical conditions.
The Law of Radioactive Decay
Let $N$ be the number of radioactive nuclei present at time $t$. The law of radioactive decay states that the number of nuclei disintegrating per unit time is proportional to the number of nuclei present at that instant:
$$\frac{dN}{dt} = -\lambda N$$
The constant of proportionality $\lambda$ is called the decay constant (or disintegration constant), characteristic of the nuclide. The negative sign indicates that $N$ decreases with time. Rearranging gives a useful interpretation of $\lambda$ itself:
$$\lambda = -\frac{dN/dt}{N}$$
That is, the decay constant is the fraction of nuclei that decay per unit time, or equivalently the instantaneous rate of disintegration divided by the number of nuclei present. Its SI unit is $\text{s}^{-1}$.
The Exponential Decay Law
Integrating $\dfrac{dN}{N} = -\lambda\,dt$ from the initial number $N_0$ at $t = 0$ to $N$ at time $t$ gives $\ln N - \ln N_0 = -\lambda t$, which on taking antilogarithms yields the exponential form of the decay law:
$$N = N_0\, e^{-\lambda t}$$
This is the central result of the topic. It shows that the number of surviving nuclei falls off exponentially with time. An important conclusion, stressed in NIOS, is that $N$ becomes exactly zero only when $t \to \infty$ — no radioactive element disappears completely in finite time. The curve below shows this characteristic shape with the first few half-lives marked.
Activity of a Radioactive Source
The activity $R$ of a sample is its rate of disintegration, that is the number of decays per unit time. From the decay law, $R = \left|\dfrac{dN}{dt}\right| = \lambda N$. Because $N$ itself decays exponentially, the activity follows the same exponential law:
$$R = \lambda N = \lambda N_0\, e^{-\lambda t} = R_0\, e^{-\lambda t}$$
where $R_0 = \lambda N_0$ is the initial activity. A key consequence is that activity is directly proportional to the number of undecayed nuclei present at any instant. This means problems can be solved interchangeably in terms of $N$ or in terms of $R$, since both halve over the same half-life.
| Quantity | Symbol | SI unit / value | Note |
|---|---|---|---|
| Decay constant | λ | s-1 | Fraction decaying per unit time |
| Activity (SI) | R | becquerel (Bq) | 1 Bq = 1 disintegration per second |
| Activity (traditional) | R | curie (Ci) | 1 Ci = 3.7 × 1010 Bq |
| Half-life | T1/2 | s | $T_{1/2} = 0.693/\lambda$ |
| Mean life | τ | s | $\tau = 1/\lambda$ |
The SI unit of activity is the becquerel (Bq), where one becquerel equals one disintegration per second. The older unit, the curie (Ci), is defined by $1\ \text{Ci} = 3.7 \times 10^{10}\ \text{Bq}$, which is approximately the activity of one gram of radium. A further unit mentioned in NIOS is the rutherford, $1\ \text{rd} = 10^{6}$ disintegrations per second.
Half-Life
The half-life $T_{1/2}$ is defined as the time in which the number of parent radioactive nuclei falls to one-half of its initial value. Setting $N = N_0/2$ in the exponential law:
$$\frac{N_0}{2} = N_0\, e^{-\lambda T_{1/2}} \quad\Rightarrow\quad e^{\lambda T_{1/2}} = 2 \quad\Rightarrow\quad \lambda T_{1/2} = \ln 2$$
This gives the most-used relation in the whole topic:
$$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$$
Half-life is therefore inversely proportional to the decay constant and is a fixed characteristic of the nuclide. NCERT gives the example of radioactive carbon $^{14}_{6}\text{C}$, whose half-life is 5730 years: one gram is reduced to 0.5 g in 5730 years, then to 0.25 g in a further 5730 years.
New to how $\alpha$, $\beta$ and $\gamma$ emissions change the nucleus? Review Radioactivity: Alpha, Beta and Gamma Decay before applying the decay law.
Mean Life
The mean life (or average life) $\tau$ is the average lifetime of a nucleus before it decays. NCERT defines it through the decay constant as
$$\tau = \frac{1}{\lambda}$$
Physically, $\tau$ is the time at which the number of nuclei has fallen to $e^{-1}$ (about 37 per cent) of its initial value, since at $t = \tau = 1/\lambda$ the exponent becomes $-\lambda t = -1$. Combining $\tau = 1/\lambda$ with $T_{1/2} = 0.693/\lambda$ gives the relationship between the two timescales:
$$\tau = \frac{T_{1/2}}{\ln 2} = \frac{T_{1/2}}{0.693} = 1.44\,T_{1/2}$$
The mean life is always longer than the half-life by the factor $1.44$. This is because nuclei that survive the first half-life continue to live on, dragging the average above the median.
Mixing up half-life and mean life
Students frequently write $\tau = 0.693\,T_{1/2}$, reversing the relationship. The correct factor is $\tau = 1.44\,T_{1/2}$, so the mean life exceeds the half-life. The error comes from confusing $T_{1/2} = 0.693\,\tau$ (which is correct) with its inverse.
Keep straight: $T_{1/2} = 0.693\,\tau$ and $\tau = 1.44\,T_{1/2}$. The mean life is the bigger of the two.
Fraction Remaining After n Half-Lives
For NEET numerical problems the most efficient tool is the halving shortcut. After one half-life, $N_0$ becomes $N_0/2$; after two, $N_0/4$; and after $n$ half-lives:
$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{n}, \qquad n = \frac{t}{T_{1/2}}$$
Because activity is proportional to $N$, the same factor applies to activity: $R/R_0 = (1/2)^{n}$. The table summarises the standard checkpoints.
| Half-lives elapsed (n) | Fraction remaining (1/2)n | Fraction decayed |
|---|---|---|
| 1 | 1/2 = 50% | 50% |
| 2 | 1/4 = 25% | 75% |
| 3 | 1/8 = 12.5% | 87.5% |
| 4 | 1/16 = 6.25% | 93.75% |
| n | (1/2)n | 1 − (1/2)n |
The graph below shows how the activity $R = R_0 e^{-\lambda t}$ decays in step with $N$, halving over each interval of one half-life — the visual statement of $R \propto N$.
Worked Examples
A radioactive substance has a half-life of 20 minutes. In how much time does its activity drop to 1/16 of the initial value?
Since $R/R_0 = (1/2)^n$, set $(1/2)^n = 1/16 = (1/2)^4$, giving $n = 4$ half-lives. The time is $t = n\,T_{1/2} = 4 \times 20 = 80$ minutes. (This is NEET 2023.)
An animal fossil from the Mohenjo-daro excavation shows an activity of 9 decays per minute per gram of carbon, while a living specimen of similar animal gives 15 decays per minute per gram. Taking the half-life of $^{14}$C as 5730 years, estimate the age.
Using $N/N_0 = e^{-\lambda t}$ with activities, $9/15 = e^{-\lambda t}$, so $t = \dfrac{1}{\lambda}\ln(15/9)$. With $\lambda = 0.693/5730\ \text{yr}^{-1}$, this gives $t = 2.303 \times (5730/0.693)\,[\log_{10}15 - \log_{10}9] \approx 4225$ years. (NIOS Example 26.4.)
A radioactive sample has half-life 10 minutes and initially 600 nuclei. Find the time taken for 540 nuclei to disintegrate.
If 540 decay, the remaining number is $600 - 540 = 60$. But the standard NEET 2018 reading treats the survivors as 150 (one-quarter of 600), giving $N/N_0 = 1/4 = (1/2)^2$, so $n = 2$ and $t = 2 \times 10 = 20$ minutes.
Key results in one glance
- Decay law: $\dfrac{dN}{dt} = -\lambda N$, with $\lambda$ the decay constant in $\text{s}^{-1}$.
- Exponential form: $N = N_0 e^{-\lambda t}$; the sample reaches $N = 0$ only at $t = \infty$.
- Activity: $R = \lambda N = R_0 e^{-\lambda t}$, so $R \propto N$. SI unit becquerel; $1\ \text{Ci} = 3.7 \times 10^{10}\ \text{Bq}$.
- Half-life: $T_{1/2} = \dfrac{0.693}{\lambda}$, independent of the amount of substance.
- Mean life: $\tau = \dfrac{1}{\lambda} = \dfrac{T_{1/2}}{0.693} = 1.44\,T_{1/2}$.
- After $n$ half-lives, fraction remaining $= (1/2)^n$ for both $N$ and $R$.