Radioactive decay and finding half life

[SUchica10]

A sample of tritium 3 decayed to 94.5% of its original amount after a year.
(a) What is the half-life of tritium-3?
(b) How long would it take the sample to decay to 20% of its original
amount?

The only equations given in my book are dm/dt = km and m(t) = m(0)e^kt but the mass is not given in the problem. Is there another equation I should be using or am I just not seeing what exactly I should be using from the problem?

 

[Kurdt]

It doesn't matter what mass it is because you know what ratios they have to decay to.

 

[arunbg]

Hint: Use the second eqn. Let initial mass be at t=0, ie m(0) in th expression. So at t = 1 year, m(t) becomes 94.5% of m(0).
Can you proceed now ?
Of course as Kurdt said you will find the final expression independent of mass, since only the ratio that remains is what matters.

 

[AlephZero]

"The sample decayed to 94.5% of it's original amout."

So if the initial mass is M, what is the mass after 1 year?

 

[PSOA]

$$m(1)=0,945m(0)$$. Do you see why?

So that m(0) now cancel out and you have:

$$0,945=e^{-k}$$

You can now find k.

 

[SUchica10]

I understand it now and I'm pretty sure I got it I just have one question... why is it e^(-k)? I understand t = 1 so that just leaves me with k but why is it negative?

 

[Kurdt]

Because it describes the decay of a radioactive substance thus the power to which an exponential should be raised to describe decay must be negative. A positive power would lead to growth.

 

[HallsofIvy]

You don't really need the exponential form. Since the question is ask for "half-life", write
$$M(t)= M_0(\frac{1}{2})^\frac{t}{T}$$
Since for t= T, that multiplies M_0 by 1/2, T is the half-life.

When t= 1,
$$M(1)= .954M_0= M_0(\frac{1}{2})^\frac{1}{T}$$
so
$$.954= (\frac{1}{2})^\frac{1}{T}$$
Take logarythms of both sides and solve for T.

 

[SUchica10]

I'm really sorry but now I'm confused.

 

[Grogs]

Your book didn't really do you any favors on the k thing. Since the amount of radioactive material remaining decreases as time increases, the first equation should properly be written as:

$$\frac{dm}{dt}=-kt$$

The minus sign on the kt represents the fact that m is decreasing as time increases (and vice-versa.)

When you solve that equation, you'll get

$$m=m_{0}e^{-kt}$$

That's good. As time increases, m decreases. k will be the decay constant, which is just ln(2)/(half-life). With the way your equations are written, k is actually the negative decay constant. It should be pretty obvious when you try to solve it, since the half-life has to be a positive value.

Anyway, remember that k isn't actually the value you're looking for in part a. If has units of inverse time. half-life = ln(2)/k is what you want.

 

[HallsofIvy]

I'm really sorry but now I'm confused.

What is it that confuses you? Do you understand that If the half life is T, then the amount is multiplied by 1/2 every T years? Do you know how to solve $$.954= (\frac{1}{2})^\frac{1}{T}$$?

As I said, take logarithms of both sides.