Question

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?

Answer 1

*bookmark

Answer 2

thanks (:

Answer 3

JAMES CAN YOU HELP <3

Answer 4

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?

Answer 5

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

Answer 6

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

Answer 7

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."

Answer 8

Hence T = ... what?

Talk to me.

Answer 9

ok so t = 4.5 billion years right?

Answer 10

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.

Answer 11

ok, i'm listening.

Answer 12

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?

Answer 13

.31/1

Answer 14

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

Answer 15

uhm, after the half life none of it.

Answer 16

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

Answer 17

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

Answer 18

uhm the ratio multiplied by t right?

Answer 19

\[ \frac{C}{C+D} \]

Therefore what fraction of the total of U + Pb is Uranium?

Answer 20

okay so itd be C / C+D

Answer 21

okay so U / U + Pb

Answer 22

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

Answer 23

honestly, i wish i could tell you

Answer 24

We just calculated Pb/U above

Answer 25

Oh!
.31/1

Answer 26

Pb/U = 0.31

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

Answer 27

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

Answer 28

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?

Answer 29

do we have to differentiate?

Answer 30

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

Answer 31

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

Answer 32

T = half life
t = elapsed time

Answer 33

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

Answer 34

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
\[ d(t) = \frac{U}{U + Pb} \]

Answer 35

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

Answer 36

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.

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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.

Answer 42

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

Answer 43

ah ok thank you so much!!!

Answer 44

What answer do you get for t?

Answer 45

i got 1.75

Answer 46

right or wrong?

Answer 47

Me too. t = 1.75 billion years.

Answer 48

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

Answer 49

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?

Answer 50

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.

Answer 51

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