After some ancient bones were excavated, carbon was taken from one of the bones to find that there was 17 % as much Superscript 14 Baseline Upper C as from current bones. The decay constant of Superscript 14 Baseline Upper C is k almost equals 0.0001216 , for time in years. How many years old is the bone?

The bone is 14572 years old.

Step-by-step explanation:

There is 17% as much Carbon-14 in the ancient bones as from current bones and the decay constant is given as 0.0001216 for time in years.

The equation for radioactive decay is N = N₀ e^ (-λt)

where t = time in years

λ = decay constant

N = final amount of carbon

N₀ = initial amount of carbon.

There is 17% as much Carbon-14 in the ancient bones as in current bones. So,

N = 17% of N₀

N = 0.17N₀

So, N/No = 0.17.

N = N₀ e^ (-λt)

N/N₀ = e^ (-λt)

0.17 = e^ (-0.0001216t)

ln (0.17) = - 0.0001216t

-0.0001216t = - 1.7719568

t = - 1.7719568/-0.0001216

t = 14572 years