The age of wine can be determined by measuring the trace amount of radioactive tritium, 3H, present in a sample. Tritium is formed from hydrogen in water vapor in the upper atmosphere by cosmic bombardment, so all naturally ocurring water contains a small amount of this isotope. Once the water is in a bottle of wine, however, the formation of additional tritium from the water is negligible, so the tritium initially present gradually diminishes by a first-order radioactive decay with a half-life of 12.5 years. If a bottle of wine is found to have a tritium concentration that is 0.168 that of freshly bottled wine (i.e. [3H]t = 0.168 [3H]0), what is the age of the wine?
To determine the age of the wine, we can use the formula for radioactive decay:
[3H]t = [3H]0 * e^(-kt)
Where [3H]t is the tritium concentration at time t, [3H]0 is the initial tritium concentration, k is the decay constant, and e is the base of the natural logarithm.
Given that [3H]t = 0.168 [3H]0, we can substitute these values into the equation:
0.168 [3H]0 = [3H]0 * e^(-kt)
Canceling out [3H]0 on both sides, we get:
0.168 = e^(-kt)
To solve for t, we need to take the natural logarithm of both sides:
ln(0.168) = ln(e^(-kt))
Using the property ln(e^x) = x, we can simplify further:
ln(0.168) = -kt
Now, we can solve for t by rearranging the equation:
t = -(ln(0.168)) / k
The half-life of tritium is 12.5 years, which means k is equal to ln(2) divided by 12.5. Substituting this value into the equation, we can find the age of the wine:
t = -(ln(0.168)) / (ln(2) / 12.5)
Calculating this expression will give us the age of the wine.
To determine the age of the wine, we can use the first-order radioactive decay equation. The equation for radioactive decay is given by:
N(t) = N(0) * e^(-kt)
Where:
N(t) = the remaining quantity of the radioactive substance at time t
N(0) = the initial quantity of the radioactive substance
k = the decay constant
t = time
In this case, we are given that the tritium concentration of the wine is 0.168 times that of freshly bottled wine. Let's assume [3H]t represents the tritium concentration at time t and [3H]0 represents the tritium concentration of freshly bottled wine.
So, we have:
[3H]t = 0.168 [3H]0
Now, the half-life of tritium is given as 12.5 years, which means after each 12.5 years, the concentration is reduced to half.
Using the decay equation, we can express the ratio of [3H]t to [3H]0 as:
[3H]t = [3H]0 * e^(-kt)
Since we are given that [3H]t is 0.168 times [3H]0, we can rewrite the equation as:
0.168 [3H]0 = [3H]0 * e^(-kt)
We can cancel out [3H]0 from both sides:
0.168 = e^(-kt)
Now, we need to find the value of kt to calculate the age of the wine. We know that the half-life (t1/2) is 12.5 years.
The relationship between the decay constant (k) and the half-life (t1/2) is given by:
k = 0.693 / t1/2
Substituting the value of t1/2 = 12.5 years:
k = 0.693 / 12.5
k ≈ 0.05544 (rounded to five decimal places)
Now, substitute the value of k back into the equation:
0.168 = e^(-0.05544t)
To find the age of the wine, we can isolate t by taking the natural logarithm (ln) of both sides:
ln(0.168) = ln(e^(-0.05544t))
Using the property of logarithms, we can simplify further:
ln(0.168) = -0.05544t * ln(e)
The natural logarithm of e (ln(e)) is equal to 1, so we have:
ln(0.168) = -0.05544t
Now, rearrange the equation to solve for t:
t = ln(0.168) / -0.05544
Using a calculator, evaluate the right-hand side of the equation:
t ≈ 36.15 years
Therefore, the age of the wine is approximately 36.15 years.
.168=e^-.693t/12.5
take ln of each side
ln(.168)=-.694t/12.5
solve for time t, in years.