A2,000g of c-14 is left to decay radioactively. the half-life of carbon-14 is approximately 5,700 years. what fraction of the sample will remain after 17,100 years?

17,100 years corresponds 3 half-lives of the carbon-14; in fact, the half-life of c-14 is 5,700 years, and we see that

after each half-life, the amount of the initial sample is halved. therefore, the amount of sample left after 3 half-lives will be:

2,000 g --> 1,000 g --> 500 g --> 250 g

which corresponds to

of the initial sample.

for this problem, we use the formula for radioactive decay which is expressed as follows:

an = aoe^-kt

where an is the amount left after time t, ao is the initial amount and k is a constant. 

we calculate as follows:

an = aoe^-kt
0.5 = e^-k(5700)
k = 1.22x10^-4

an = aoe^-kt
an = 2000 e^-1.22x10^-4(17100)
an = 248 g left