Question

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?

Answer 1

Answer:

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

Answer 2

After ancient bones are excavated the carbon was taken and analyzed form one for the bone and had 17& as superscripts 14 baseline and upper C as from the current bone.

• 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.

• For radioactive decay is N = N₀ e^(-λt) There is 17% as much Carbon-14 in the ancient bones as in current bones. So,

• N = 17% of N₀

• N = 0.17N₀.

Lean more about the ancient bones.

brainly.com/question/14953752.