radioactive exponential decay

Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.


Experiments show that at each instant a radioactive substance decomposes—and is thus decaying in time—proportional to the amount of substance present.



放射性物質の量、たとえば0.5 g(グラム)が与えられたら、後で存在する量を見つけます。





Step 1. Setting up a mathematical model of the physical process. Denote by y(t) the amount of substance still present at any time t.
By the physical law , the time rate of change y'(t) = dy/dt is proportional to y (t) this gives the first-order ODE.

(6) dy/dt = -kt

where the constant k is positive , so that , because of the minus , we do get decay.
The value of k is known from experiments for various radioactive substances (e.g., k=1.4/10^-11sec^-1 approximately , for radium )

ステップ1.物理プロセスの数学モデルを設定します。いつでも存在する物質の量をy(t)で示します。 物理法則により、時間変化率y ‘(t)= dy / dtはy(t)に比例し、これにより1次ODEが得られます。 dy / dt = -kt ここで、定数kは正であるため、マイナスのため、減衰します。 kの値は、さまざまな放射性物質の実験から知られています(例:k = 1.4 / 10 ^ -11sec ^ -1、ラジウムの場合)

Now the given initial amount is 0.5 g, and we can call the corresponding instant Then we have the
initial condition This is the instant at which our observation of the process begins. It motivates
the term initial condition (which, however, is also used when the independent variable is not time or when
we choose a t other than ). Hence the mathematical model of the physical process is the initial value problem

与えられた初期量は0.5 gであり、対応するインスタントを呼び出すことができます

(7) dy/dt = -kt , y(0) = 0.5.

Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decay
and has the general solution (with arbitrary constant c but definite given k)


(8) y(t) = ce^-kt

We now determine c by using the initial condition. Since y(0) = c from (8), this gives y(0) = c = 0.5. Hence
the particular solution governing our process is (cf. Fig. 5)

(9) y(t) = 0.5 e^-kt

ここで、初期条件を使用してcを決定します。 (8)からy(0)= cであるため、これによりy(0)= c = 0.5が得られます。したがって

Always check your result—it may involve human or computer errors! Verify by differentiation (chain rule!)
that your solution (9) satisfies (7) as well as y(0) = 0.5:

dy/dt = -0.5ke^-kt = -k/0.5e^-kt = -ky , y(0) = 0.5e^0 = 0.5.

Step 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts from
the correct initial amount and decreases with time because k is positive. The limit of y as t → ∞  is zero.

正しい初期量であり、kが正であるため時間とともに減少します。 t→∞としてのyの制限はゼロです。




放射性指数減衰 (https://ja.wikipedia.org/wiki/%E6%8C%87%E6%95%B0%E9%96%A2%E6%95%B0%E7%9A%84%E6%B8%9B%E8%A1%B0)