Respuesta :
a) One glass has the fresh wine; the one with the higher tritium concentration. To determine the age of the other one, we have to resort to a trick; we notice that 64 is 2 raised to the power 6. Hence, 6 half-lifes need to have passed so that the concentration of tritium becomes 1/64 of the initial concentration (every year the concentration halves, so after n years it is [tex] (\frac{1}{2} )^n[/tex]). Thus, the age of the old wine is approximately 6*12.29 =73.5 years.
b) This is not an exact science. The major assumption that underlies this method is that initially each wine had the same tritium count. This can just be flat out wrong, since one variety of grapes could have been grown in soil that is more rich in tritium. Secondly, we need to be sure that the wine was not stored near tritum etc. It is also important to standardize for wine size; double a quantity of wine would give off double the titrium count, even if it were the same age. One can make further comments like that, but the basis of radiochronology (timing with radioactivity) is to know whether the two samples (one of known age, one of unknown) had the same concentration of radioactive materials initially or what that ratio of concentrations was; in this case we needed to assume that they were the same to do our calculation.
b) This is not an exact science. The major assumption that underlies this method is that initially each wine had the same tritium count. This can just be flat out wrong, since one variety of grapes could have been grown in soil that is more rich in tritium. Secondly, we need to be sure that the wine was not stored near tritum etc. It is also important to standardize for wine size; double a quantity of wine would give off double the titrium count, even if it were the same age. One can make further comments like that, but the basis of radiochronology (timing with radioactivity) is to know whether the two samples (one of known age, one of unknown) had the same concentration of radioactive materials initially or what that ratio of concentrations was; in this case we needed to assume that they were the same to do our calculation.