Having studied the data for patterns and trends, you should now be able to draw some conclusions about knowing the half-life of the element Lokium would help you determine the absolute age of rock in which this element is found. imagine you have found a rock that has 5 grams (a.k.a. cubes) of Lokium and 95 grams of DOL. Determine the absolute age of that rock. Assume that each trail in your experiment represents 1,000 years. Show all work.
I guess you have a rock weighing 100 grams
I guess it started out all Lokium at t = 0 ?
if it has a half life of T years
then
fraction Lokium = (1/2)^n where n is the number of half lives
here
fraction of Lolium seems to be 5/100
so
5/100 = (1/2)^n
log .05 = n log .5
n = log .05/log .5 = -1.3/ -0.301 = 4.33 half lives
so that rock is 4.33 half lives old
Now I guess you have experimental data that gives you fraction of Lokium versus time
something like'
fraction mass/original mass = f = e^-kt
find k from the data using ln f = - k t
then for half life we want to solve for T when the fraction is 1/2
1/2 = e^-kT
ln 0.5 = - k T
T = 0.693 / k where T is the half life
could you do the fraction Lolium be 40/100
To determine the absolute age of the rock, we need to use the concept of radioactive decay and the half-life of the element Lokium.
Given that we have a rock with 5 grams of Lokium and 95 grams of DOL, we need to determine how much Lokium would have been present initially.
Let's assume the half-life of Lokium is 10,000 years.
Initially, let's assume there were 100 grams of Lokium and 0 grams of DOL in the rock.
After the first 10,000 years, half of the Lokium would have decayed. So, we would have 50 grams of Lokium and 50 grams of DOL.
After the next 10,000 years, half of the remaining 50 grams of Lokium would have decayed. So, we would have 25 grams of Lokium and 75 grams of DOL.
Continuing this pattern, we can calculate the amount of Lokium and DOL at each time interval:
- After 20,000 years: 12.5 grams of Lokium and 87.5 grams of DOL.
- After 30,000 years: 6.25 grams of Lokium and 93.75 grams of DOL.
- After 40,000 years: 3.125 grams of Lokium and 96.875 grams of DOL.
Since we started with 100 grams of Lokium, after 40,000 years, only 3.125 grams of Lokium remain in the rock.
The absolute age of the rock would be 40,000 years since that is the time it took for the Lokium to decay to 3.125 grams.
Note that this method assumes a constant decay rate for Lokium and a closed system (no additional element or isotope entering or leaving the rock).
To determine the absolute age of the rock, we need to use the concept of half-life. Half-life is the time it takes for half of a radioactive substance to decay.
Unfortunately, the question doesn't provide the half-life of Lokium. So, without that information, it is impossible to calculate the absolute age of the rock accurately.
However, I can explain the general concept of how you would calculate the absolute age if you knew the half-life.
Let's assume that the half-life of Lokium is X number of trails, where each trail represents 1,000 years. Here's how you can calculate the absolute age of the rock:
1. Start with the initial amount of Lokium in the rock, which is 5 grams.
2. After X trails, half of the Lokium will have decayed, leaving you with 2.5 grams of Lokium.
3. After another X trails (for a total of 2X trails), another half of the remaining Lokium will have decayed. You'll be left with 1.25 grams of Lokium.
4. Repeat this process until you reach a point where the remaining Lokium is negligible (let's say less than 0.005 grams, for example).
5. Count the number of X trails required to reach this negligible amount.
6. Multiply the number of X trails by 1000 years to calculate the absolute age of the rock.
Keep in mind that this is a hypothetical explanation since the specific half-life of Lokium is not provided in the question.