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?
12th grade as the school subject, doesn't help much. Please say Math, Science, whatever it may be to get help from the correct volunteer teacher.
Sra
To calculate the decay constant of ^14C, we can use the formula:
decay constant (λ) = ln(2) / half-life
Here, ln(2) is the natural logarithm of 2.
Plugging in the given value of the half-life (5730 years), we can calculate the decay constant as follows:
λ = ln(2) / 5730
Now let's calculate it:
λ = 0.693 / 5730 ≈ 0.000120968
Therefore, the decay constant of ^14C is approximately 0.000120968 per year.
To determine the number of ^14C nuclei remaining after 22920 years, we can use the formula for exponential decay:
N(t) = N₀ * e^(-λt)
Here, N₀ represents the initial number of nuclei, e is the base of the natural logarithm (approximately 2.71828), λ is the decay constant, and t is the time in years.
Let's calculate the number of ^14C nuclei remaining after 22920 years, assuming we started with 1000 nuclei:
N(22920) = 1000 * e^(-0.000120968 * 22920)
Now, let's do the calculations:
N(22920) ≈ 1000 * e^(-2.77)
Using a calculator, we find:
N(22920) ≈ 1000 * 0.0633 ≈ 63.3
Therefore, after 22920 years, approximately 63.3 ^14C nuclei will be remaining from the original sample of 1000 nuclei.