| English | 意味 |
|---|---|
| random/ˈrændəm/ | ランダム |
| spontaneous/spɒnˈteɪnɪəs/ | 自発的 |
| count rate/kaʊnt reɪt/ | カウントレート |
| activity/ækˈtɪvɪti/ | 放射能 |
| decay constant/dɪˈkeɪ ˈkɒnstənt/ | 崩壊定数 |
| radionuclide/ˌreɪdiəʊˈnjuːklaɪd/ | 放射性核種 |
| becquerel/ˈbekwərəl/ | ベクレル |
| half-life/hɑːf laɪf/ | 半減期 |
| exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ | 指数関数的減衰 |
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Activity = decay constant × number of undecayed nuclei, measured in becquerel.
Match each quantity to its definition.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$A = \lambda N = 0.010 \times 1.0 \times 10^{6} = 1.0 \times 10^{4}\ \text{Bq}$.
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.
半減期 — 原子核の減衰を見よう
各原子核にはランダムな減衰の一定確率があります。時間を進めると、約半数の原子核が減衰し、半減期ごとに残りの数が半分になり、さらに半分になります。
The activity of a radioactive source decays exponentially with time.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
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:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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 $\lambda = 0.693/t_{1/2}$