Question

A bottle of vintage red wine has lost its label. The concentration of a radioactive isotope of hydrogen (hydrogen-3 or tritium) in the wine is 0.34 times that found in freshly bottled wines. If the half-life of tritium is 12.3 years, calculate the time elapsed since the wine was bottled. This is a first order decay.

Answer 1

To calculate the time elapsed since the wine was bottled, we can use the formula for first-order decay. The concentration of a radioactive isotope decreases by half every half-life. In this case, the concentration of tritium in the wine is 0.34 times that found in freshly bottled wines. Using the half-life of tritium and the given concentration ratio, we can determine the time elapsed.

Step-by-step explanation:

To calculate the time elapsed since the wine was bottled, we can use the formula for first-order decay. The concentration of a radioactive isotope decreases by half every half-life. In this case, the concentration of tritium in the wine is 0.34 times that found in freshly bottled wines. Since tritium has a half-life of 12.3 years, we can calculate the time elapsed as follows:

t = (ln(Cf/Co))/(-k)

Where t is the time elapsed, Co is the initial concentration, Cf is the final concentration, and k is the decay constant. Since Co = 1 and Cf = 0.34, we can substitute these values into the formula and solve for t:

t = (ln(0.34))/(-k)

Now we need to find k. The decay constant can be determined from the half-life using the equation:

k = (ln(2))/t_half

Substituting the value for t_half = 12.3 years, we can calculate k:

k = (ln(2))/(12.3)

Finally, substituting the value of k into the first equation, we can solve for t:

t = (ln(0.34))/(-(ln(2))/(12.3))

t = (12.3)(ln(0.34))/ln(2)

Answer 2

19 years

Step-by-step explanation:

Tritium follows a first-order decay which can be represented by the following expression.

`ln(([H]_(t))/([H]_(0)) )=-k.t`

where,

[H]t is the concentration of tritium at a certain time t

[H]₀ is the initial concentration of tritium

k is the rate constant

If we know the half-life (t1/2), we can calculate the rate constant.

`k=(ln2)/(t_(1/2)) =(ln2)/(12.3y) =0.0564y^(-1)`

`ln(([H]_(t))/([H]_(0)) )=-k.t\\ln((0.34[H]_(0))/([H]_(0)) )=-(0.0564y^(-1)).t\\t=19y`