You have before you two glasses of Amontillado wine- one made last year and one made many years ago.
Tritium undergoes radioactive decay with a half-life of 12.33
years.The glass on the left has sixty-four times fewer tritium counts than the glass on the right.
a. What is the age of the wine in the glass on the left, and the glass on the right?
b. What assumption(s) do you have to make in order to compare their tritium counts?
c. Is this an accurate assumption in light of the events of the post- World War II world?
So far, I’ve figured out the rate to be k= 0.005621 but I’m lost on what to do after that
✅ Answers
? Best Answer
For simple radioactive decay the arte of change of radioactive atoms is given by [1]:
dN/dt = – λ∙N
where λ is the decay constant and N the number of radioactive atoms∙
When you solve this differential subject to initial condition
N(t=0) = N₀
you get the following relation for the number of radioactive particles at time t.
N(t) = N₀∙e^{ – λ∙t }
Half-life t_½ and decay constant are related as:
λ = ln(2)/t½
You can derive this relation from:
(1/2)∙N₀ = N(t½) = N₀∙e^{ – λ∙t½ }
The counter doesn’t measure the number of atoms. It measures the activity of the sample, i.e. the number of radioactive decays per unit time, which is proportional to the number of radioactive atoms:
A = – dN/dt = λ∙N = λ∙N₀∙e^{ – λ∙t }
Assume that both samples of wine contained the same number of atoms at the year of the production. Then counts for each glass are given by:
– left glass
A₁ = λ∙N₀∙e^{ – λ∙t₁ }
– right glass
A₂ = λ∙N₀∙e^{ – λ∙t₂ }
Since the exponential function decreases with time t, The gals with higher count, i.e. the right glass, must be the younger wine. Therefore
t₂ = 1 a
When you consider the ratio of the activities the unknown N₀ cancels out:
A₂/A₁ = [λ∙N₀∙e^{ – λ∙t₂ }] / [λ∙N₀∙e^{ – λ∙t₁ }]
= e^{ – λ∙t₂ }/e^{ – λ∙t₁ }
= e^{ – λ∙t₂ } ∙ e^{ λ∙t₁ }
= e^{ λ∙(t₁ – t₂ } }
<=>
ln(A₂/A₁) = λ∙(t₁ – t₂ }
<=>
t₁ = t₂ + ( ln(A₂/A₁)/λ )
<=>
t₁ = t₂ + ( t_½ ∙ ln(A₂/A₁)/ln(2) )
The activity of the right glass is 64 times higher i.e.
A₂/A₁ = 64
Hence,
t₁ = 1 y + ( 12.33 a ∙ ln(64)/ln(2) ) = 74.98 y ≈ 75 a
That means the wine is 75 years old, its from 1938.
b.
Basic assumption is that the environment (soil, air, water etc) of the Amontillado vineyard contains a constant concentration of tritium – more precisely: a constant fraction of the hydrogen atoms in the environment are present as isotope tritium. The flourishing grape-vines are in equilibrium with their environment and take the material, which they need to grow and to form fruit, from their environment. So their ripe grapes have the content of tritium as the environment. At the instant, when the grapes are harvested and converted to wine the content of tritium is on that level too. But then the wine (in the barrel or in the bottle) is no longer in equilibrium with its environment. So the the tritium content decreases due to radioactive decay.
c:
It’s inaccurate (of course 😉 ).
After second world war (primarily in the fifties) there were a lot of atmospheric nuclear weapon tests, which caused a substantial rise of radioactive isotopes in the earth atmosphere. This increased radioactivity in the atmosphere was the reason for the Limited Test Ban Treaty[2] from 1963, which prohibits all nuclear test except underground test. In the wikipedia article you find a graph showing the atmospheric content C14 (which also used for dating). A plot for tritium would look similar but withe a narrower peak, because tritium decays faster than C14 with half-life 5700 a.
However you can expect the initial tritium content in 2012 to be greater than before the 50s.
That’s why the older Amontillado is younger than 75 years.
.