Half-Life problem...need help fast...

[elisemd]

So I have a half-life problem that I do not know how to solve. It goes somewhat like this: The half-life of Radium 226 is about 1620 years. How much time is needed for a sample weighing 2.0g to decay to 0.1g?

I thought the equation went like this:
f(t) = f(0)e^kt.

Now which numbers are supposed to go where???

I would greatly appreciate it if I got some help on this. Thank you!!
~Elise[/b]

 

[Unco]

G'day, Elise.

If we are to use that equation, set t=0 when the mass of Radium 226 is 2.0g.

That is, f(0) = 2.0, and we have:

. . . \(\displaystyle \L f(t) \, = \, 2e^{(kt)}\)

An half-life of 1620 years means its takes 1620 years for the mass of Radium 226 to halve.

So after 1620 years, that is when t=1620, the mass of Radium 226 will be 1g (half of 2g).

Substitute f(1620) = 1 into the equation:

. . . \(\displaystyle \L 1 \, = \, 2e^{(1620k)}\)

Divide through by 2:

. . . \(\displaystyle \L \frac{1}{2} \, = \, e^{(1620k)}\)

Take natural logs of both sides:

. . . \(\displaystyle \L \ln{\left(\frac{1}{2}\right)} \, = \, 1620k\)

Solve for k:

. . . \(\displaystyle \L k = \frac{\, \ln{\frac{1}{2}} \, }{ \, 1620 \, } \, \approx \, -0.0004279\)

The formula becomes
. . . \(\displaystyle \L f(t) \, = \, 2e^{(-0.0004279t)}\)

To find the time need for the sample to have mass 0.1g, substitute f(t)=0.1 into the formula and solve for t.

You can check your answer using a more intuitive formula for radioactive decay:

. . . \(\displaystyle \L m_{final} \, = \, m_{initial} \, \times \, \left(\frac{1}{2}\right)^n\)

Where \(\displaystyle n\) is the number of decays to have occurred; given by times elapsed / half-life.

 

[galactus]

Hint:

\(\displaystyle k=\frac{-1}{T}ln(2)=\frac{-1}{1620}ln(2)=-.00042786863\)

\(\displaystyle A_{0}e^{kt}=2e^{kt}\)

\(\displaystyle 2e^{kt}=0.1\)

You know k, solve for t.