The radioactive isotope of Carcon (^14C)has a half life of 5730. What is the decay constant of ^14C? If we start with a sample of 1000 carcon nuclei, how many will be left in 22920 years time?

22920 years is exactly 4 half-lives, so the number of nuclei is 2^4 = 16 times less. That will leave 1000/16 = 67 C14 atoms.

The relationship of the decay constant (k) and the half-life (T1/2) is:

k = 0.693/(T1/2)

so k = 1.21*10^-4 years-1

To find the decay constant of a radioactive isotope, we can use the formula:

λ = ln(2) / T½

Where:
λ is the decay constant
ln is the natural logarithm
2 is the constant for half-life
T½ is the half-life of the isotope

Plugging in the given values for ^14C with a half-life of 5730 years, we can calculate the decay constant:

λ = ln(2) / 5730 ≈ 0.000120968

Now, let's calculate the number of ^14C nuclei that will be left in 22920 years, using the formula:

N(t) = N₀ * e^(-λt)

Where:
N(t) is the number of nuclei at time t
N₀ is the initial number of nuclei
e is the mathematical constant approximately equal to 2.71828 (the base of the natural logarithm)
λ is the decay constant
t is the time period

Plugging in the values for N₀ = 1000 (initial number of nuclei), λ = 0.000120968 (decay constant), and t = 22920 years:

N(t) = 1000 * e^(-0.000120968 * 22920)

Using a calculator, we can compute the value:

N(t) ≈ 455.71

Therefore, approximately 455 ^14C nuclei will be left after 22920 years.