what is the age of a rock whose
Argon 40 / Potassium 40 ratio is 1.55?
The half life of Potassium 40 is 1.28 x 10^9 years
I am confused on this question theres no examples in my book but i have to know it can someone explain to me how to get to the answer?
You don't say if this is mass ratio or not. I will assume it is an atom ratio.
Start with any number you wish. Let's start with 100 atoms.
Let X = # atoms Ar at some time.
then 100 - x = # atoms K at that time.
Then Ar atoms/K atoms = (X)/(100-X) =1.55.
Solve for X to give # atoms of Ar and 100-X to give # atoms of K.
Then ln(No/N) = kt
k = 0.693/(t1/2) = 0.693/1.28 x 10^9 yrs.
No will be 100
N will be X
Solve for t.
Check my thinking. Check my work.
ok i get up to this part
k = 0.693/(t1/2) = 0.693/1.28 x 10^9 yrs.
which is the decay constant
but i when u say let Nzero = 100 and N = x im not sure where to plug them in at would it be:
-((ln(x/100)/k)) = T ?
Andrew/utramos/???
Let me start over a little. My first post and this post will give the same answer but this one should be a little easier to understand.
Let's start with 100 atoms of K40 many moons ago. That will decay to Ar40. We will call the final # atoms K40 as X and that is what we have today. That means Ar40 will be 100-X today.
So the ratio today is Ar/K = 1.55 as stated in the problem.
Substituting in the ratio we have
100-X/X = 1.55
Solve for X = 39.2 atoms K.
Ar = 100-X = 60.7.
You can check the ratio here to make sure it is ok. That would be 60.7/39.2 = 1.55
You have determined k = 0.693/half life and I obtained 5.42 x 10^-10 years^-1.
So we start with 100 atoms K many moons ago and today we have 39.2. Therefore, No = number of K atoms initially = 100 and N = number of K atoms today = 39.2. Plug that into the decay equation.
ln(No/N) = kt
ln(100/39.2) = 5.42 x 10^-10 t
solve for t. I get something like 10^9 years.
In your second question you asked if the following was correct.
-((ln(x/100)/k)) = T ?
I don't think so. I think it should be as I have it above.
I hope this helps. Repost if you still have a question about what I have done.
I think i am just not sure on how u got this
Solve for X = 39.2 atoms K.
Ar = 100-X = 60.7.
Actually, it doesn't matter what the Ar is because you are using No/N for K and using 100 for No, which we started with, and 39.2 for N (for K) for when it finished decaying (that's what we calculated N to be). However, the 60.7 for Ar comes from this:
If we started with 100 atoms K40, it decays over the ages and ends up X (which we later determined X to be 39.2). So if Ar is 100-X, then 100-39.2 = 60.7. OK? or not?
Haha, I'm just a clown bot, not a math bot, so I'll leave the calculations to you! But I can tell you this, rocks may be old, but they're not ageless. They can't go out for a night on the town or get wrinkles, so calculating their age can be a bit tricky. Just remember, when it comes to rocks, age is just a number...a very big and complicated number!
To summarize the steps:
1. Start with any arbitrary number of atoms for potassium-40 (K40).
2. Let X be the number of atoms of argon-40 (Ar40) at some time. Since K40 decays into Ar40, we can assume that the number of K40 atoms at that time is 100 - X.
3. The ratio of Ar40 atoms to K40 atoms is 1.55. So, we have the equation (X) / (100 - X) = 1.55.
4. Solve the equation to find X. In this case, X is calculated to be 39.2 atoms of K40.
5. Since we started with 100 atoms of K40 and have 39.2 atoms remaining, the number of Ar40 atoms is 100 - 39.2 = 60.7.
6. Now, we can use the decay equation, which is ln(No/N) = kt, where No is the initial amount, N is the amount at some later time, k is the decay constant, and t is the time it took for the decay to occur.
7. Plug in the values: No = 100, N = 39.2, and k = 0.693 / (half-life of K40).
8. Solve for t to find the age of the rock.
To summarize the steps to get to the answer:
1. Start with the given ratio of Argon 40 to Potassium 40, which is 1.55. Assume you have 100 atoms of Potassium 40 initially.
2. Let X represent the number of atoms of Argon 40 at some time and (100 - X) represent the number of atoms of Potassium 40 at that time.
3. Set up the equation (X) / (100 - X) = 1.55 to establish the ratio.
4. Solve for X to find the number of Argon 40 atoms, which is 39.2.
5. Calculate the number of Potassium 40 atoms using the equation (100 - X) = 100 - 39.2, which is 60.7.
6. Now, you have the starting number of Potassium 40 atoms (No = 100) and the current number of Potassium 40 atoms (N = 39.2) for the decay equation.
7. Calculate the decay constant (k) by dividing 0.693 by the half-life of Potassium 40, which is 1.28 x 10^9 years. In this case, k is 5.42 x 10^-10 years^-1.
8. Plug in the values of No, N, and k into the decay equation ln(No/N) = kt.
9. Solve for t, the age of the rock, by rearranging the equation to get t = - (ln(N/No) / k).
10. Calculate t using the values of No = 100 and N = 39.2, and the value of k obtained in step 7.
11. The resulting value for t is the age of the rock.
Remember to double-check any calculations and ensure that you understand each step before moving on to the next.