Word Problem with Constant 'e'

R=?No, you don't put in 5700 for t. That would be the answer if the ratio was 0.5/10^12You are given that the ratio is 1/10^13 and you want to find t.So start by putting in 1/10^13 for R and solve for t.So, t would be 11,400?Good: so if the ratio is 0.5/10^12, the creature died 5700 years ago. If the ratio is 1/10^13, then it died 11400 years ago. And if the ratio is 1/10^14, then it died 17100 years ago.
  • #1
darshanpatel
139
0

Homework Statement



For living organic material, the ratio of the number of radioactive carbon isotopes (carbon-14) to the number of non-radioactive carbon isotopes (carbon-12) is about 1 to 10^12. When organic material dies, its carbon 12 content remains fixed whereas its radioactive carbon-14 bagins to decay with a half-life of about 5700 years. To estimate the age of dead organic material, scientists use the following formula which denotes the ratio of carbon-14 to carbon-12 present at any time 't' (in years).

R=(1/10^12)e^(-t/8223)

Estimate the age of newly discovered fossil in which the ratio of carbon-14 to carbon-12 is 1/10^13

Homework Equations



-None-

The Attempt at a Solution



I just substituted one thing, I got confused, do I plug in 5,700 for 'R' or 11,400 for 'R' then solve?

From this: R=(1/10^12)e^(-t/8223)
To this: R=(1/10^13)e^(-t/8223)

That is as far as I got

Please help! Thanks
 
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  • #2
You have to understand what the different terms mean.
The general equation would be:
[tex]R=R_0e^{-t/\tau}[/tex] where
[itex]R_0[/itex] is the ratio that it started out with (at t=0);
[itex]\tau[/itex] is the "mean-life" of the process in question, related to the half-life[1] by [itex]\tau[/itex]=T1/2/ln(2); and
[itex]R[/itex] is the ratio now (i.e. after some time t has passed).

eg. for C14, the half-life is 5700yrs so the mean life is 8223yrs. So [itex]\tau=8223\text{yrs}[/itex].

So - is 1/10^13 the ratio it started with or the ratio now?
Where does it go?

----------------------------
note: Also written R=R0eλt
... which should look familiar if you ever played "hλlf life" :)

λ = ln(2)/T1/2
 
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  • #3
Yeah I don't understand any of what you just said. I know you replace the 1/10^12 to 1/10^13 because they gave you a new ratio. But for what everything is equal to, 'R,' do I put in 5700 or what?
 
  • #4
Yeah I don't understand any of what you just said.
If you don't at least try you are going to fail your course!
I know you replace the 1/10^12 to 1/10^13 because they gave you a new ratio.
This is not correct.
But for what everything is equal to, 'R,' do I put in 5700 or what?
The answer is "what". Please be aware that I am not going to do your homework for you. You have to do it yourself.
So you have to figure out for yourself what to put for R, but I can guide you.
You figure it out by understanding what it means - what does R, in the equation stand for? What does it mean?
What kind of thing is the 5700years? What is it called?

Consider what the equation is telling you:
If the creature had died today, then the ratio would be 1/10^12
In 5700 years the ratio would have been 0.5/10^12 because e^(5700/8223)=0.5

This also means that if the creature had dies 5700 years ago, the ratio back then would have been 1/10^12 and now it will read as 0.5/10^12.

Get it yet?
 
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  • #5
Oh, I think I get it.. So for the thing to decay fully we plug 11400 into 't' because it said it began to decay with a half-life of about 5700 years.

Once we plug it in, we can put it in the calculator and found out what 'R' is right?

I am saying I do this:

R=(1/10^13)e^(-11400/8223)
?
 
  • #6
Please answer the questions.
What does R stand for in the equation?
What does t stand for in the equation?

What is the quantity you are trying to find?

Also please answer the questions in post #4.
Those are not rhetorical - they are put there for you to tell me the answers.
 
  • #7
What I said in thread starter post, is all the problem gives me, that's why I am confused just like you.. Sorry.

All I can answer for you is that 't' stands for time in years only because that is given.
 
  • #8
Time is correct for t.

The answer to everything I asked you is in post #1 and post #2.

Do you even know what the point of the equation is?
My problem is that in order to guide you, I need to find some point in your course where you do understand something. But, if I take what you have said at face value, you don't even understand basic algebra. I'll see if I can get someone more experienced to see if they can figure out what to do.
 
  • #9
Can you please show me what the formula would look like with everything substituted in? Maybe then I might understand
 
  • #10
Nice try: that sort of thing has a special name, it is called "doing your homework for you". We don't do that :)

I did give you some examples in post #4
Consider what the equation is telling you:
If the creature had died today, then the ratio would be 1/10^12
In 5700 years the ratio would have been 0.5/10^12 because e^(5700/8223)=0.5

This also means that if the creature had died 5700 years ago, the ratio back then would have been 1/10^12 and now it will read as 0.5/10^12.

OK - so let's do the last one:
A body is found that has a C14 ratio of 0.5/10^12
How long ago did it die?

We reason that if the current C14 ratio is R, then t will be the time since the C14 ratio was at maximum. (post #2) We write[tex]\frac{0.5}{10^{12}} = \frac{1}{10^{12}}e^{t/8332}[/tex]then solve for t.
 
  • #11
darshanpatel said:
To estimate the age of dead organic material, scientists use the following formula which denotes the ratio of carbon-14 to carbon-12 present at any time 't' (in years).

R=(1/10^12)e^(-t/8223)
Okay, they are saying that R:

(1) is equal to the expression above,

and

(2) is the ratio of carbon-14 to carbon-12 present at any time 't' ...

Estimate the age of newly discovered fossil in which the ratio of carbon-14 to carbon-12 is 1/10^13
And here they are telling us that at some time 't', the ratio of carbon-14 to carbon-12 is equal to 1/1013.

In other words:

R is the ratio of carbon-14 to carbon-12,​
and
the ratio of carbon-14 to carbon-12 is 1/1013.​

Therefore, R = _____?
 
  • #12
would i put 5700 in for 't' and change 1/10^12 to 1/10^13?

Then solve for R?
 
  • #13
darshanpatel said:
would i put 5700 in for 't' and change 1/10^12 to 1/10^13?

Then solve for R?
No.

There is a simple, logical connection you have to make here:
Redbelly98 said:
R is the ratio of carbon-14 to carbon-12,​
and
the ratio of carbon-14 to carbon-12 is 1/1013.​

Therefore, R = _____?
Fill in the blank, use this value for R, then solve for t.
 

Related to Word Problem with Constant 'e'

What is the constant 'e' in a word problem?

The constant 'e' is a mathematical constant that is approximately equal to 2.71828. It is also known as Euler's number and is a fundamental constant in calculus and other areas of mathematics.

How is the constant 'e' used in word problems?

The constant 'e' is commonly used in word problems involving exponential growth or decay, such as compound interest or population growth. It is also used in problems involving natural logarithms and exponential functions.

How do I solve a word problem with the constant 'e'?

To solve a word problem with the constant 'e', you will need to identify the relevant information and use the appropriate formula or equation. This may involve setting up an exponential or logarithmic equation and using basic algebraic techniques to solve for the unknown variable.

What is the significance of the constant 'e' in mathematics?

The constant 'e' has many important applications in mathematics, including in calculus, complex analysis, and number theory. It is also closely related to the imaginary unit 'i' and plays a key role in the study of exponential and logarithmic functions.

Are there any real-life applications of the constant 'e'?

Yes, the constant 'e' has numerous real-life applications, particularly in the fields of finance, biology, and physics. For example, it is used to calculate compound interest, model population growth, and describe the behavior of radioactive decay.

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