- anonymous

Somebody please help meee!!!
Suppose that U^238 has a half life of 4.5 billion years, decaying (through a series of
relatively short lived intermediate atoms) to Pb^206. In a certain mineral sample there
are .31 times as many Pb^206 atoms as there are of U^238. If one assumes that the
mineral deposit contained no Pb^206 when it was formed and that no lead or uranium
have been added to or escaped from the sample (except through the natural decay
process) how old is the sample?

- schrodinger

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- JamesJ

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- anonymous

thanks (:

- anonymous

JAMES CAN YOU HELP <3

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## More answers

- JamesJ

Right. Having a half life of time T means that after time t, a material will be
\[ d(t) = 2^{-t/T} \]
of what it was originally.
For example at time t= 0, it should be all itself, so
\[ d(0) = 2^0 = 1.\]
At time t= T, it should have half decayed, so it's half of what it was,
\[ d(T) = 2^{-T/T} = 2^{-1} = \frac{1}{2} \]
After time t=2T, it should be half of half of the original substance, i.e., 1/4. And that's what our formula gives us, as
\[ d(2T) = 2^{-2T/T} = 2^{-2} = \frac{1}{4}. \]
So are you convinced this formula d(t) makes sense?

- anonymous

yes the formula makes sense, i just dont understand how i'm suppose to use it with lead and uranium together ..

- JamesJ

Ok. In this problem, what is the half time, T?

- JamesJ

It's right there at the beginning of your problem:
"Suppose that U^238 has a half life of 4.5 billion years, decaying (through a series of
relatively short lived intermediate atoms) to Pb^206."

- JamesJ

Hence T = ... what?
Talk to me.

- anonymous

ok so t = 4.5 billion years right?

- JamesJ

The half life T = 4,500,000,000 years, yes. What we want to find is the time t. So let's be careful to differentiate between little t and capital T.

- anonymous

ok, i'm listening.

- JamesJ

Now, let U denote the amount of Uranium and let Pb denote the amount of iron. What do we know about the ratio:
\[ \frac{Pb}{U} \] What is it equal to?

- anonymous

.31/1

- JamesJ

Right. Now, in terms of U and Pb, what fraction of the original U is still Uranium?

- anonymous

uhm, after the half life none of it.

- JamesJ

No. The total amount of material now at time t is U + Pb. Therefore what fraction of the total is Uranium?

- JamesJ

If there C cats and D dogs, what fraction of all the animals are cats?

- anonymous

uhm the ratio multiplied by t right?

- JamesJ

\[ \frac{C}{C+D} \]
Therefore what fraction of the total of U + Pb is Uranium?

- anonymous

okay so itd be C / C+D

- anonymous

okay so U / U + Pb

- JamesJ

Right, and
\[ \frac{U}{U + Pb} = \frac{1}{1 + Pb/U} \]
Now, what's that equal to?

- anonymous

honestly, i wish i could tell you

- JamesJ

We just calculated Pb/U above

- anonymous

Oh!
.31/1

- JamesJ

Pb/U = 0.31
Therefore
\[ \frac{U}{U+Pb} = \frac{1}{1+Pb/U} = \frac{1}{1.31} \]
Yes?

- anonymous

ok ya that makes sense.. sorry i'm such a retard btw aha

- JamesJ

Now, what we have is that
\[ d(t) = \frac{U}{U + Pb} \]
You have a formula for the left-hand side (LHS) and now you know the value of the RHS. What formula do you now have then for t?

- anonymous

do we have to differentiate?

- JamesJ

Nope. \[ d(t) = 2^{-t/T} \] remember.

- anonymous

ok so that is equal to what you said above,
whats the different between the big t and the little t?

- JamesJ

T = half life
t = elapsed time

- anonymous

because one of them is 4.5 billion years right? and then we just have to solve for the other one?
but i dont understand what we sub in for the different elements then

- JamesJ

We have after a time t, \( d(t) \) of the Uranium is still Uranium. We need to solve for t. Now, another expression for the amount of material that is still Uranium is
\[ \frac{U}{U + Pb} \]
because Pb is what the U decays into.
Therefore if we know the value of \( \frac{U}{U + Pb} \) we can find the time t that has passed in order that
\[ d(t) = \frac{U}{U + Pb} \]

- anonymous

and we do know that, so we can solve for t?

- JamesJ

Now that last equation is equivalent to
\[ 2^{-t/T} = \frac{1}{1.31} \]
You know the value of the half life T. Now solve for elapsed time t.

- anonymous

ok, when solving should i put the whole 4.5 billion years in or just keep it at 4.5?

- JamesJ

You can use whatever units you like. Time units of billion years is perfectly good.

- anonymous

so then if i solve for t as 4.5 that should be fine as long as my answer is in billion years?

- JamesJ

T = 4.5. NOT, NOT t. You're solving for t.

- anonymous

ok right.
can i ask one more question?
how would i go about solving for the variable
i know that if its e with an exponent you just ln it, but i'm not sure if its an actual number.

- JamesJ

\[ 2^{-t/T} = 1/1.31 \]
take the log base 2 of both sides and you have
\[ \frac{-t}{T} \log_2 2 = \log_2 (1/1.31) \]

- anonymous

ah ok thank you so much!!!

- JamesJ

What answer do you get for t?

- anonymous

i got 1.75

- anonymous

right or wrong?

- JamesJ

Me too. t = 1.75 billion years.

- anonymous

oh yay i did it right!! ok thank you so much!

- anonymous

i know i've already asked you a billion questions because your so smart, but do you know how to change basis of transformation matricies?

- JamesJ

Let T be your original matrix with respect to a basis E = {e1, e2, ... , en}. Let F = {f1, f2, ..., fn} be another basis and U the matrix that transforms E to F. Then if you want T in basis F, then you want to use
\[ UTU^{-1} \]
I think the logic for this is clear. \( U^{-1} \) takes a vector in basis F, puts it in terms of the basis of E. \( T \) then operates on it. Then \( U \) transforms that back into the basis F.

- anonymous

ok thanks! i'm trying to get something from its basis in F, back to the standard basis!

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