KAHIBARO
Table of Contents
The mathematical form of radioactive decay
A radioactive sample contains many unstable nuclei. Each unstable nucleus has some chance to decay, and over time the number of undecayed nuclei decreases. The important point is that the decrease does not happen by losing the same number of nuclei in equal time intervals. Instead, the sample loses the same fraction of its remaining nuclei in equal time intervals. This behavior is called exponential decay.
If $N(t)$ is the number of undecayed nuclei at time $t$, then the decay law is
$$
\frac{dN}{dt} = -\lambda N
$$
where $\lambda$ is the decay constant. The minus sign shows that $N$ decreases with time.
The defining rule of exponential decay is
$$
\frac{dN}{dt} = -\lambda N
$$
This means the rate of decay is proportional to the number of undecayed nuclei still present.
Solving the decay equation
To find how $N$ changes with time, we solve the differential equation. Separating variables gives
$$
\frac{dN}{N} = -\lambda \, dt
$$
Integrating both sides,
$$
\int \frac{dN}{N} = \int -\lambda \, dt
$$
so
$$
\ln N = -\lambda t + C
$$
where $C$ is a constant. If at $t = 0$ the number of undecayed nuclei is $N_0$, then
$$
N(0) = N_0
$$
and the solution becomes
$$
N(t) = N_0 e^{-\lambda t}
$$
This is the exponential decay law.
The exponential decay formula is
$$
N(t) = N_0 e^{-\lambda t}
$$
where $N_0$ is the initial number of radioactive nuclei and $\lambda$ is the decay constant.
What the formula means
The factor $e^{-\lambda t}$ tells us what fraction of the original nuclei remain after time $t$. As time increases, this factor becomes smaller and smaller, but it never becomes exactly zero.
For example, if after some time the exponent is $-1$, then
$$
N = N_0 e^{-1} \approx 0.368 N_0
$$
So about $36.8\%$ of the original nuclei remain.
If the exponent is $-2$, then
$$
N = N_0 e^{-2} \approx 0.135 N_0
$$
So about $13.5\%$ remain.
This shows that exponential decay is fast at first, when many nuclei are present, and slower later, when fewer are left.
Fraction remaining and fraction decayed
Sometimes we do not need the exact number of nuclei, only the fraction remaining:
$$
\frac{N(t)}{N_0} = e^{-\lambda t}
$$
The fraction that has decayed by time $t$ is
$$
1 - \frac{N(t)}{N_0} = 1 - e^{-\lambda t}
$$
These forms are often useful in practical calculations.
Useful forms of the decay law are
$$
\frac{N(t)}{N_0} = e^{-\lambda t}
$$
and
$$
\text{fraction decayed} = 1 - e^{-\lambda t}
$$
Shape of the decay curve
A graph of $N$ versus $t$ starts at $N_0$ when $t=0$, then drops smoothly and continuously. The curve is steep at the beginning and becomes flatter with time.
This curve is different from a straight line. In linear decrease, the same amount would disappear every second. In radioactive decay, the amount disappearing each second gets smaller because fewer undecayed nuclei remain.
Equal fractions in equal times
One of the most important features of exponential decay is that equal time intervals remove equal fractions, not equal numbers.
Suppose we look at the sample every time interval $\Delta t$. Then
$$
N(t+\Delta t) = N(t)e^{-\lambda \Delta t}
$$
So each time interval multiplies the remaining number by the same factor.
For example, if $e^{-\lambda \Delta t} = 0.80$, then every interval leaves $80\%$ of what was there before. That means $20\%$ decays in each interval, regardless of how many nuclei remain.
This explains why radioactive decay is called a constant fractional process.
Units of the decay constant
Since the exponent in an exponential must be dimensionless, the quantity $\lambda t$ must have no units. Therefore $\lambda$ must have units of inverse time.
If time is measured in seconds, then $\lambda$ has units of $\text{s}^{-1}$. If time is measured in years, then $\lambda$ can be written in $\text{year}^{-1}$.
The larger $\lambda$ is, the faster the sample decays.
| Decay constant $\lambda$ | Meaning |
|---|---|
| Small $\lambda$ | Slow decay |
| Large $\lambda$ | Fast decay |
The decay constant $\lambda$ has units of inverse time, such as $\text{s}^{-1}$.
A larger $\lambda$ means a more rapid decrease in $N(t)$.
Example calculation
Suppose a sample begins with
$$
N_0 = 1000
$$
radioactive nuclei, and its decay constant is
$$
\lambda = 0.20 \ \text{day}^{-1}
$$
How many undecayed nuclei remain after $5$ days?
Use
$$
N(t) = N_0 e^{-\lambda t}
$$
Substitute the values:
$$
N(5) = 1000 e^{-(0.20)(5)} = 1000 e^{-1}
$$
Since $e^{-1} \approx 0.368$,
$$
N(5) \approx 1000 \times 0.368 = 368
$$
So about $368$ nuclei remain.
Comparing exponential and linear decrease
It is helpful to compare exponential decay with a straight line decrease.
| Type of decrease | Mathematical form | Key idea |
|---|---|---|
| Linear decrease | $N = N_0 - kt$ | Same amount lost per unit time |
| Exponential decay | $N = N_0 e^{-\lambda t}$ | Same fraction lost per unit time |
Radioactive decay follows the exponential form, not the linear one.
A microscopic view
At the level of a single nucleus, we cannot predict exactly when it will decay. But for a very large number of nuclei, the overall behavior becomes very regular. The random decays combine to produce the smooth exponential law.
This is why a radioactive sample can be unpredictable for one nucleus, but very predictable for a huge collection of nuclei.
A useful graphical interpretation
If we plot $N(t)$ directly against $t$, we get a curved graph. But if we take the natural logarithm of the decay law,
$$
N(t) = N_0 e^{-\lambda t}
$$
then
$$
\ln N = \ln N_0 - \lambda t
$$
This has the form of a straight line,
$$
y = b + mx
$$
with slope $-\lambda$ and intercept $\ln N_0$.
So a plot of $\ln N$ versus $t$ is a straight line. This is very useful when analyzing experimental decay data.
Taking the natural logarithm of the decay law gives
$$
\ln N = \ln N_0 - \lambda t
$$
A graph of $\ln N$ versus $t$ is a straight line with slope $-\lambda$.
Summary of the decay law
Exponential decay describes how the number of undecayed radioactive nuclei decreases with time. The key idea is that the decay rate is proportional to the number still present. This leads to the differential equation
$$
\frac{dN}{dt} = -\lambda N
$$
and its solution
$$
N(t) = N_0 e^{-\lambda t}
$$
The quantity $\lambda$ controls how quickly the decay happens. Equal time intervals remove equal fractions of the sample, and a plot of $\ln N$ versus $t$ is a straight line. This law is one of the most important mathematical descriptions in nuclear physics.