Radioactive decay
| English | Chinese | Pinyin |
|---|---|---|
| random | 随机 | suí jī |
| spontaneous | 自发 | zì fā |
| count rate | 计数率 | jì shù lǜ |
| activity | 活度 | huó dù |
| decay constant | 衰变常数 | shuāi biàn cháng shù |
| radionuclide | 放射性核素 | fàng shè xìng hé sù |
| becquerel | 贝克勒尔 | bèi kè lēi ěr |
| half-life | 半衰期 | bàn shuāi qī |
| exponential decay | 指数衰减 | zhǐ shù shuāi jiǎn |
Clicks that never come evenly
- Put a Geiger counter beside a long-lived source and listen. The clicks come in bursts and gaps, never in a steady rhythm.
- Count for ten seconds, again and again, and no two counts match. Yet the long-run average is perfectly steady.
- That fluctuation is not sloppy apparatus. It is the evidence that decay is random 随机.
- This lesson is what "random and spontaneous" means precisely, and the exponential law that follows from it.
Random and spontaneous
- Spontaneous 自发 means the decay happens with no outside trigger, and the rate is unaffected by temperature, pressure or chemical state.
- Random means it is impossible to predict which nucleus will decay next, or when a given nucleus will decay. Only a probability can be given.
- The evidence for randomness is the fluctuating count rate: counts in equal intervals from the same source are never the same twice, even when the activity is not changing.
- Naming the two words is not the answer. The marks are on not affected by external factors and cannot predict which or when.
A steady average made of unsteady counts
Select all the true statements about radioactive decay.
It is spontaneous and random, and the rate does not depend on conditions. You can only give the probability of a single decay.
What is the evidence that radioactive decay is random?
A falling activity shows decay happens; a constant half-life shows the probability is fixed; heating shows it is SPONTANEOUS. Only the fluctuation shows randomness.
Activity and the decay constant
- For $N$ undecayed nuclei of a radionuclide 放射性核素:
- Activity 活度 is the number of decays per unit time, measured in becquerel 贝克勒尔 (Bq), where $1\ \text{Bq} = 1$ decay per second.
- The decay constant 衰变常数 is the probability per unit time that a nucleus will decay, also in $\text{s}^{-1}$.
- They share a unit but they are not the same idea: $A$ counts events, $\lambda$ is a probability. Calling $\lambda$ "the rate of decay" is the standard lost mark.
The activity of a radioactive source is:
Activity = decay constant × number of undecayed nuclei, measured in becquerel.
Match each quantity to its definition.
Activity and the decay constant share the unit s^-1 but are different quantities. Calling the decay constant "the rate of decay" is the standard lost mark.
Worked example: a tiny mass, a huge activity
- Fluorine-18 has a half-life of $110$ minutes. Show that $\lambda = 1.05\times10^{-4}\ \text{s}^{-1}$, and find the activity of $2.1\times10^{-12}\ \text{kg}$ of it.
- $\lambda = \dfrac{0.693}{110\times60} = 1.05\times10^{-4}\ \text{s}^{-1}$. Convert the half-life to seconds first, or $\lambda$ comes out per minute.
- Number of nuclei: $N = \dfrac{2.1\times10^{-12}}{18\times1.661\times10^{-27}} = 7.0\times10^{13}$.
- $A = \lambda N = (1.05\times10^{-4})(7.0\times10^{13}) = 7.4\times10^{9}\ \text{Bq}$.
- A picogram gives billions of decays a second because the half-life is short. The same mass of uranium-238, half-life $4.5\times10^{9}$ years, gives about $10^{-5}\ \text{Bq}$.
A source has $\lambda = 0.010$ per second and $1.0 \times 10^{6}$ undecayed nuclei. What is its activity?
$A = \lambda N = 0.010 \times 1.0 \times 10^{6} = 1.0 \times 10^{4}\ \text{Bq}$.
Why the decay is exponential
- Since $\lambda$ is the fractional decay rate, $\dfrac{dN}{dt} = -\lambda N$, and the solution is an exponential decay 指数衰减:
- Any count rate 计数率 proportional to the activity follows the same law, which is why the equation is usually quoted as $x = x_0 e^{-\lambda t}$.
- The three-mark explanation: (1) $\lambda$ is the probability per unit time and is the same for every nucleus whatever its age; (2) so the rate of decay is proportional to the number of undecayed nuclei, $A = \lambda N$; (3) a rate of change proportional to the quantity itself gives an exponential, meaning the same fraction decays in each equal interval.
- So the number never quite reaches zero. It only halves, again and again.
Decay equations (α, β, γ)
Choose a decay type; the daughter nuclide is fixed so the nucleon number A and the proton number Z both balance.
Half-life — watch the nuclei decay
Each nucleus has a fixed chance of decaying, at random. Move time forward: about half the remaining nuclei decay every half-life — so the count halves, then halves again.
The activity of a radioactive source decays exponentially with time.
The same fraction decays each second, giving $A = A_0 e^{-\lambda t}$.
Put the three-mark explanation of why decay is exponential in order.
Each of the first three is a mark. Writing only the formula N = N0 e^(-lambda t) answers none of them.
Half-life
- The half-life 半衰期 $t_{1/2}$ is the time for the number of undecayed nuclei, or the activity, or the count rate, to fall to half.
- Put $N = N_0/2$ into the exponential law and the $N_0$ cancels:
- Because $N_0$ cancels, the half-life does not depend on how much you started with. A gram and a tonne of the same nuclide halve in the same time.
The half-life is related to the decay constant by:
Setting $N = \tfrac{1}{2}N_0$ in $N = N_0 e^{-\lambda t}$ gives $\lambda t_{1/2} = \ln 2$.
The half-life of a nuclide is unaffected by which of these? Select all that apply.
N0 cancels in the half-life derivation, so the amount is irrelevant, and spontaneity rules out external conditions. The nuclide itself is the ONLY thing that sets it.
Worked example: reading a decay backwards
- A source's activity falls from $8.0\times10^{5}\ \text{Bq}$ to $1.0\times10^{5}\ \text{Bq}$ in $24$ minutes. Find the half-life and the decay constant.
- The activity has fallen to $\tfrac{1}{8}$, and $\tfrac{1}{8} = (\tfrac{1}{2})^3$, so three half-lives have passed: $t_{1/2} = 24/3 = 8.0$ minutes.
- $\lambda = \dfrac{0.693}{8.0\times60} = 1.4\times10^{-3}\ \text{s}^{-1}$.
- When the ratio is a neat power of $\tfrac{1}{2}$, count half-lives. Otherwise take logs: $\lambda t = \ln(A_0/A)$.
- Always convert the half-life to seconds before quoting $\lambda$ in $\text{s}^{-1}$.
What fraction of the nuclei remain after 3 half-lives?
Each half-life halves the number: $\left(\tfrac{1}{2}\right)^{3} = \tfrac{1}{8} = 0.125$.
A source's activity falls from 8.0e5 Bq to 1.0e5 Bq in 24 minutes. What is its half-life, in minutes?
The activity is one eighth, and 1/8 = (1/2)^3, so three half-lives fit in 24 minutes. When the ratio is not a neat power of a half, take logs instead: lambda t = ln(A0/A).
Before using a measured count rate in a decay calculation, subtract the ____ count.
The background is quoted in the question precisely so that you subtract it. Forgetting to makes every derived half-life too long.
Marks that slip away
- Activity is decays per unit time. The decay constant is a probability per unit time. Same unit, different quantity.
- The evidence for randomness is the fluctuating count rate, not the fact that the activity falls.
- Convert the half-life to seconds before using $\lambda = 0.693/t_{1/2}$.
- Half-life is independent of the amount of substance, and of temperature, pressure and chemical state.
- $x_0$ in $x = x_0 e^{-\lambda t}$ is the value at $t = 0$, not the value at the start of the observation, unless the clock is set there.
- Subtract the background count before using a measured count rate. It is quoted in the question precisely so you will.
You've got it
- decay is spontaneous (unaffected by external factors) and random (which nucleus and when are unpredictable), evidenced by a fluctuating count rate
- activity is decays per unit time in becquerel, the decay constant is the probability of decay per unit time, and $A = \lambda N$
- decay is exponential because a fixed probability per nucleus makes the rate proportional to the number left: $x = x_0 e^{-\lambda t}$
- the half-life is the time to fall to half, is independent of the starting amount, and $\lambda = 0.693/t_{1/2}$