carbon dating

[jokerman]

2. Suppose you have 500 grams of a radioactive element who half-life is 27 years.
a) Find the function for the quantity, Q, left in t years.

b) Find to the nearest tenth of a year when there will be 100 grams left.

i asked this in a different thread and someone tried to help me but lost me completely, could somebody walk me through this problem from beginning to finish so i can see how its done.

the equation given in my book for this kind of problem is: P(t) =P0e^-kt , k > 0 (P0 is a P with a small zero ahead of and below the P). it says p0 is the quantity present at time 0, P(t) is the amount present at time t, and k is the decay rate. now im sure that all make sense if you've seen a problem like this before but i havn't so if anybody could walk me through step by step using that equation to get the answers to a and b that would be a great help. thanks

 

[galactus]

The P_0 means the amount you start out with. In this case 500 grams.

In 27 years it reduces by 1/2. In other words, there will be 250 grams left in 27 years.

\(\displaystyle \L\\250=500e^{-k(27)}\)

\(\displaystyle \L\\\frac{1}{2}=e^{-27k}\)

\(\displaystyle \L\\ln(\frac{1}{2})=-27k\)

\(\displaystyle \L\\\frac{ln(\frac{1}{2})}{-27}=\frac{ln(2)}{27}=k\)

There's k. Now, you can use the formula to find the amount at anytime t.

When will there be 100 grams left?. You have k. Set the formula equal to 100 and solve for t.

 

[soroban]

Hello, jokerman!

2. Suppose you have 500 grams of a radioactive element who half-life is 27 years.

a) Find the function for the quantity, \(\displaystyle Q(t)\), left in \(\displaystyle t\) years.

b) Find to the nearest tenth of a year when there will be 100 grams left.

The half-life formula looks like this: \(\displaystyle \:Q(t) \:=\:Q_oe^{-kt}\)

where \(\displaystyle Q_o\) is the original amount of the substance
. . and \(\displaystyle k\) is a constant to be determined.

The formula for \(\displaystyle k\) is: \(\displaystyle \:k\:=\:\frac{\ln(2)}{\text{half-life}}\;\) *


For our problem: \(\displaystyle \:Q_o \,=\,500\,\) and \(\displaystyle \,k \:=\:\frac{\ln(2)}{27} \:\approx\:0.02567\)

(a) Hence, the function is: \(\displaystyle \L\:\fbox{Q(t) \:=\:500e^{-0.02567t}}\)


When is \(\displaystyle Q(t)\,=\,100\) ?

We have: \(\displaystyle \:500e^{-0.02567t} \:=\:100\;\;\Rightarrow\;\;e^{-0.02567t} \:=\:0.2\)

Take logs: \(\displaystyle \:\ln\left(e^{-0.02567t}\right) \:=\:\ln(0.2)\;\;\Rightarrow\;\;-0.02567t\cdot\ln(e) \:=\:\ln(0.2)\)

(b) Therefore: \(\displaystyle \:t \;=\;\frac{\ln(0.2)}{-0.02567} \;=\;62.69723071 \;\approx\;\L\fbox{62.7\text{ years}}\)

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* .If you'd like an explanation of that \(\displaystyle k\)-formula, let me know.

Edit: I see that Galactus explained most of it already . . .

 

[jokerman]

Thank you so much!
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Edited by stapel -- Reason for edit: Removing obscenities.