# Homework Help: Questions about the Differential Equation (Dacay)

1. Dec 6, 2009

### mybingbinghk

The half-life of a radioactive isotope is the amount of time it takes for a quantity
of radioactive material to decay to one-half of its original amount.
i) The half-life of Carbon 14 (C-14) is 5230 years. Determine the decay-rate
parameter  for C − 14.
ii)The half-life of Iodine 131 (I-131) is 8 days. Determine the decay-rate param-
eter for I − 131.

2. Relevant equations

dx/dt = -bx where b is a constant

3. The attempt at a solution

i) let x(t) be the quantity of radionactuve meterial at time t in year and dx/dt be the rate of change of the quantity.

dx/dt = -bx where b is a constant

I dun no how to process after doing this assumption. Please help to give some hints on doing these questions. Many thanks!

2. Dec 6, 2009

### foxjwill

Let's let $t_{1/2}=5230\,\text{yr}$ be the half-life of C-14. What does this mean? Well, if at some initial time $t_0$ the total quantity of C-14 is $x_0=x(t_0)$, then after an additional time $t_{1/2}$ the amount of carbon remaining is $x_0/2$; algebraically, $x(t_0+t_{1/2})=x_0/2$.

The decay constant $\lambda$, instead of specifying how long it takes for the quantity to halve, tells you the rate of change of the percent change in quantity. That's a mouthful. So, what does that mean? Well, consider the same quantities $x_0$ and $t_0$ as before. How much carbon is left after an infinitesimal time $dt$? Let's go take a look at our (verbal) definition for the decay constant: "percent change in quantity" means $dx/x(t)$ (remember that $dx=x(t+dt)-x(t)$), and the "rate of change" of this is $[dx/x(t)]/dt$. Thus, the decay constant is given algebraically by
$$\lambda=-\frac{dx}{dt}\frac{1}{x(t)}.$$​

Can you think of a way of incorporating these to find $\lambda$ for C-14?

3. Dec 6, 2009

### mybingbinghk

t0 =0, x0 = 0
When t = 5230, then I dun no how to find the X=???
then can't find the lambda?

4. Dec 6, 2009

### HallsofIvy

Starting with dx/dt = -bx, "separate" the variables by separating the derivative into differentials: dx= -bx dt so dx/x= -bdt. Integrate both sides: $\int dx/x= -b\int dt$.

5. Dec 6, 2009

### mybingbinghk

that means lambda is dx/dt, right?

As u said, I do the calculation and find the follow
ln|x| = -bt +C where C is a constant

then what should I do?

Can u explain more details?

6. Dec 6, 2009

### foxjwill

Why do you think that x0=0?

7. Dec 7, 2009

### mybingbinghk

I think if the time is zero, there is no decay so x is equal to ZERO

8. Dec 7, 2009

### foxjwill

Ahah! There's your problem. The value x measures the quantity remaining, not the quantity which has been removed.