Half life question help?

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BubbleLover98
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#1
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#1
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.

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SH0405
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#2
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#2
(Original post by BubbleLover98)
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
\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).
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BubbleLover98
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#3
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#3
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.
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SH0405
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#4
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#4
(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)

\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.
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BubbleLover98
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#5
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#5
Thank you soo much!
(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)

\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.
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SH0405
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#6
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#6
(Original post by BubbleLover98)
Thank you soo much!
No worries.
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Plagioclase
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#7
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#7
(Original post by SH0405)
 = 1.2 \times 10^{10} \ y
That's a very old rock!!
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SH0405
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#8
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#8
(Original post by Plagioclase)
That's a very old rock!!
Over 85% the age of the universe.... Life goes on. For the rock, anyway.
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Plagioclase
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#9
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#9
(Original post by SH0405)
Over 85% the age of the universe.... 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!
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SH0405
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#10
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#10
(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.
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Plagioclase
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#11
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#11
(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.
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SH0405
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#12
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#12
(Original post by Plagioclase)
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.
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.
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Plagioclase
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#13
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#13
(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.
Gneiss puns. Hopefully OP doesn't take your help for granite with these tuff questions. I breccia they couldn't gedrite without your help.
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SH0405
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#14
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#14
(Original post by Plagioclase)
Gneiss puns. Hopefully OP doesn't take your help for granite with these tuff questions. I breccia they couldn't gedrite without your help.
Hopefully OP will be Cummingtonite to join in with all the puns.
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SH0405
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#15
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#15
(Original post by SH0405)
Hopefully OP will be Cummingtonite to join in with all the puns.
(Original post by Plagioclase)
Gneiss puns. Hopefully OP doesn't take your help for granite with these tuff questions. I breccia they couldn't gedrite without your help.
A bit far-fetched perhaps.

(There's no pun here, so don't look.)
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Plagioclase
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#16
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#16
(Original post by SH0405)
Hopefully OP will be Cummingtonite to join in with all the puns.
I was waiting for you to come up with that one, it's a load of schist.
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