Question

The water in a lake contains a certain amount of tritium `sf(""^3T)`. The water is used to make wine and whisky. Given that the half - life of tritium is 12.3 years can you answer the following ?

(a) Water from the lake is made to make a certain vintage of wine. It was bottled and laid down for 30 years. What is the proportion of the tritium in the wine sample compared to that of the original source ?

(b) In a sample of whisky the proportion of `sf(""^3T)` has fallen to 3/5 of the original amount in the lake. How old is the whisky ?

Answer 1

`(a)`

`""^3T_("wine"):""^3T_("lake")=0.184`

`(b)`

`9.07" ""yr"`

Explanation:

`(a)`

The expression for 1st order radioactive decay is:

`N_(t)=N_(0)e^(-lambdat)`

`N_0` is the initial number of undecayed atoms of `""^3T`.

`N_(t)` is the number of undecayed atoms of `""^3T` after time `t` has elapsed.

`lambda` is the decay constant.

The decay constant and the 1/2 life are related by the following expression:

`lambda=0.693/t_(1/2)`

`:.lambda=0.693/12.3=0.0563" "a^(-1)`

I will assume that once the wine is bottled the amount of `""^3T` falls as it decays whereas the proportion of `""^3T` in the lake remains constant as the water is replenished.

Taking natural logs of the decay expression:

`lnN_t=lnN_0-lambdat`

`:.lnN_0-lnN_t=lambdat`

`:.ln[N_0/N_t]=lambdat`

Putting in the numbers:

`ln[N_0/N_t]=0.0563xx30=1.69`

From which

`N_0/N_t=5.42`

`:.N_t/N_0=0.1845`

This gives us the ratio of tritium in the wine to that of the lake i.e:

`""^3T_("wine"):""^3T_("lake")=0.184`

`(b)`

In the whisky the proportion of `""^3T` has fallen to 3/5 its proportion in the local source.

`:.N_t=3/5N_0`

`:.N_0/N_t=5/3`

`:.ln[5/3]=0.0563t=0.51`

`:.t=0.51/0.0563=9.07" ""yr"`

nb Scotch is "whisky"

Irish is "whiskey" and was invented by them.