Half life question help?

Uranium decays into lead. the half life of uranium is 4,000,000,000 years. A sample of radioactive rock contains 7 times as much lead as it does uranium. Calculate the age of the sample.

thanks :smile:

Reply 1

SH0405

Original post by BubbleLover98

\lambda = \frac{ln2}{t_{\frac{_1}{^2}}} = \frac{ln2}{4 \times 10^9} y^{-1}

x = x_0e^{-\lambda t}

From the given information we deduce that the sample is currently

\frac{1}{8} \ uranium

and so

e^{-\lambda t} = \frac{1}{8}

\Rightarrow -\lambda t = ln\frac{1}{8}

\Rightarrow \lambda t = ln8

\Rightarrow t = \frac{1}{\lambda}ln8

=\frac{4 \times 10^9 \times \ ln8}{ln2}

= 1.2 \times 10^{10} \ y

Not really supposed to give full solutions, but it was so tempting.

Hope that helped (and that it's correct).

(edited 8 years ago)

Reply 2

BubbleLover98

Thank you.. But Is there any other simpler way?.. I doubt they'd make us do that 2 weeks into the course lol

Original post by SH0405

\lambda = \frac{ln2}{t_{\frac{1}{2}}} = \frac{ln2}{4 \times 10^9} y^{-1}

x = x_0e^{-\lambda t}

From the given information we deduce that the sample is currently

\frac{1}{8} \ uranium

and so

e^{-\lambda t} = \frac{1}{8}

\Rightarrow -\lambda t = ln\frac{1}{8}

\Rightarrow \lambda t = ln8

\Rightarrow t = \frac{1}{\lambda}ln8

=\frac{4 \times 10^9 \times \ ln8}{ln2}

\approx 1.2 \times 10^{10} y

Not really supposed to give full solutions, but it was so tempting.

Reply 3

SH0405

Original post by BubbleLover98

Thank you.. But Is there any other simpler way?

Certainly.

You can calculate the number of half lives that have elapsed by the point at which one eighth of the sample is uranium and work from there, as shown below:

\left(\frac{1}{2}\right)^n = \frac{1}{8}

\Rightarrow n = 3

(by inspection, or calculation if truly necessary)

Unparseable latex formula:

\Rightarrow t = 3t_{\frac{_1}{^2}}}

= 3 \times 4 \times 10^{9} \ y

= 1.2 \times 10^{10} \ y

It's interesting to note how both ways I have shown you relate to one another to lead to the same answer.

(edited 8 years ago)

Reply 4

BubbleLover98

Thank you soo much! :smile:

Original post by SH0405

Certainly.

You can calculate the number of half lives that have elapsed to the point at which one eighth of the sample is uranium and work from there, as shown below:

\left(\frac{1}{2}\right)^n = \frac{1}{8}

\Rightarrow n = 3

(by inspection, or calculation if truly necessary)

Unparseable latex formula:

\Rightarrow t = 3t_{\frac{_1}{^2}}}

= 3 \times 4 \times 10^{9} \ y

= 1.2 \times 10^{10} \ y

It's interesting to note how both ways I have shown you relate to one another to lead to the same answer.

Reply 5

SH0405

Original post by BubbleLover98

Thank you soo much! :smile:

No worries. :smile:

Reply 6

Plagioclase

Original post by SH0405

= 1.2 \times 10^{10} \ y

That's a very old rock!!

Reply 7

SH0405

Original post by Plagioclase

That's a very old rock!!

Over 85% the age of the universe.... :redface:

Life goes on. For the rock, anyway.

Reply 8

Plagioclase

Original post by SH0405

Over 85% the age of the universe.... :redface:

Life goes on. For the rock, anyway.

I'm going to assume that there was a lot of lead in the rock to start with, otherwise that rock comes from outside the solar system!

Reply 9

SH0405

Original post by Plagioclase

I'm going to assume that there was a lot of lead in the rock to start with, otherwise that rock probably comes from outside the galaxy!

Hmm....not sure. I hope my maths is correct.

Reply 10

Plagioclase

Original post by SH0405

Hmm....not sure. I hope my maths is correct.

No, your maths is right. I've seen questions similar to this followed by "The earth is only 4.5 billion years old, explain the discrepancy between this and the age of the rock" and the answer is that there was lead in the rock to start with, it's quite a common thing that people have to deal with when radiodating things.

Reply 11

SH0405

Original post by Plagioclase

Well, you're solid as a rock when it comes to radioactivity knowledge. It must have taken decaydes to become that good. I'm glad I lead BubbleLover98 to the correct answer.

Reply 12

Plagioclase

Original post by SH0405

Well, you're solid as a rock when it comes to radioactivity knowledge. It must have taken decaydes to become that good. I'm glad I lead BubbleLover98 to the correct answer.