Physics · Nuclei

Law of Radioactive Decay and Half-Life

Radioactive decay is a statistical nuclear process governed by a single, elegantly simple law: the rate at which an unstable sample disintegrates is proportional to the number of nuclei still present. From this one statement, set out in NCERT Class 12 Physics §13.6 and NIOS §26.3, follow the exponential decay law, the concepts of half-life and mean life, and the activity of a source. For NEET, this cluster of formulae is among the most reliably examined in the Nuclei chapter.

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.

NEET Trap

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.

N t N₀ N₀/2 N₀/4 N₀/8 2T½ 3T½
Exponential decay of $N = N_0 e^{-\lambda t}$. In each successive half-life $T_{1/2}$ the population halves: $N_0 \to N_0/2 \to N_0/4 \to N_0/8$, approaching but never reaching zero.

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.

QuantitySymbolSI unit / valueNote
Decay constantλs-1Fraction decaying per unit time
Activity (SI)Rbecquerel (Bq)1 Bq = 1 disintegration per second
Activity (traditional)Rcurie (Ci)1 Ci = 3.7 × 1010 Bq
Half-lifeT1/2s$T_{1/2} = 0.693/\lambda$
Mean lifeτs$\tau = 1/\lambda$

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.

Build the foundation

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.

NEET Trap

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)nFraction decayed
11/2 = 50%50%
21/4 = 25%75%
31/8 = 12.5%87.5%
41/16 = 6.25%93.75%
n(1/2)n1 − (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 $R \propto N$.

R t R₀ R₀/2 R₀/4 2T½ 3T½ 4T½ R₀/4
Activity $R = R_0 e^{-\lambda t}$ falls by a factor of two every half-life. After four half-lives the activity is $R_0/16$ — the basis of the NEET 2023 question.

Worked Examples

Example 1

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.)

Example 2

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.)

Example 3

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.

Quick Recap

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 $R$.

NEET PYQ Snapshot — Law of Radioactive Decay and Half-Life

Real NEET previous-year questions on the decay law, half-life, mean life and fraction remaining.

NEET 2023

The half-life of a radioactive substance is 20 minutes. In how much time will the activity of the substance drop to 1/16 of its initial value?

  • (1) 80 minutes
  • (2) 20 minutes
  • (3) 40 minutes
  • (4) 60 minutes
Answer: (1) 80 minutes

$R/R_0 = (1/2)^n = 1/16 = (1/2)^4 \Rightarrow n = 4$. So $t = n\,T_{1/2} = 4 \times 20 = 80$ minutes.

NEET 2021

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:

  • (1) 2/3
  • (2) 1/2
  • (3) 1/(2√2)
  • (4) 2/(3√2)
Answer: (3) 1/(2√2)

$R/R_0 = (1/2)^{t/T_{1/2}} = (1/2)^{150/100} = (1/2)^{3/2} = \dfrac{1}{2\sqrt{2}}$.

NEET 2018

For a radioactive material, the half-life is 10 minutes. If initially there are 600 nuclei, the time taken (in minutes) for the disintegration of 540 nuclei is:

  • (1) 20
  • (2) 10
  • (3) 30
  • (4) 15
Answer: (1) 20

Following the official key, the remaining nuclei are taken as 150, i.e. one-quarter of 600. Thus $N/N_0 = (1/2)^2$, so $n = 2$ and $t = 2 \times 10 = 20$ minutes.

NEET 2017

Radioactive material A has decay constant 8λ and material B has decay constant λ. Initially they have the same number of nuclei. After what time will the ratio of the number of nuclei of B to that of A be 1/e?

  • (1) 1/(9λ)
  • (2) 1/λ
  • (3) 1/(7λ)
  • (4) 1/(8λ)
Answer: (3) 1/(7λ) — officially marked Bonus

$N_A = N_0 e^{-8\lambda t}$, $N_B = N_0 e^{-\lambda t}$, so $N_B/N_A = e^{7\lambda t}$. Setting this equal to $1/e$ gives $7\lambda t = -1$, a negative time, so the question was declared a bonus; the intended magnitude is $t = 1/(7\lambda)$.

FAQs — Law of Radioactive Decay and Half-Life

Quick answers to the points most often confused in NEET.

Does the half-life of a radioactive sample depend on the amount of substance present?

No. The half-life depends only on the decay constant of the nuclide through the relation T1/2 = 0.693/lambda. The decay constant is a characteristic property of the radioactive nucleus and is independent of the amount of substance, temperature, pressure or chemical state. Whether you start with one gram or one kilogram, exactly half of the nuclei present decay in one half-life.

What is the difference between half-life and mean life?

Half-life T1/2 is the time in which the number of parent nuclei falls to one-half of its initial value, and equals 0.693/lambda. Mean life tau is the average lifetime of a nucleus and equals 1/lambda, the time in which the number falls to e^(-1) (about 37 percent) of the initial value. They are related by tau = T1/2 / 0.693, that is tau = 1.44 T1/2, so the mean life is always longer than the half-life.

What fraction of a sample remains after n half-lives?

After n half-lives the fraction of the original nuclei remaining is (1/2)^n. For example, after 4 half-lives the fraction remaining is (1/2)^4 = 1/16. Since activity is proportional to the number of nuclei, the activity also drops to (1/2)^n of its initial value after n half-lives.

What are the SI and traditional units of activity?

The SI unit of activity is the becquerel (Bq), where 1 becquerel equals 1 disintegration per second. The traditional unit is the curie (Ci), defined so that 1 Ci = 3.7 x 10^10 Bq, which is approximately the activity of one gram of radium. The decay constant lambda itself is measured in per second.

Why does a radioactive sample never disappear completely?

The exponential law N = N0 e^(-lambda t) shows that N becomes exactly zero only when t is infinite. Each half-life removes only half of the nuclei present, so the number keeps halving without ever reaching zero in finite time. In practice the count eventually becomes negligible, but mathematically the decay law predicts that the sample never vanishes completely.