Carbon-14 has a half-life of 5730 y. How much of a 144 g sample of a carbon-14 will remain after 1.719 (times) 10 4y?
See your other radioisotope problem. Most of these follow the same procedure; this one is a little different but uses the same formulas. Post with explanation if you can't get it.
To find out how much of the carbon-14 sample remains after the given time, we can use the formula for exponential decay:
N = N₀ * e^(-kt)
Where:
N = the amount of carbon-14 remaining after time t
N₀ = the initial amount of carbon-14 (144 g, in this case)
e = the base of the natural logarithm (approximately 2.71828)
k = the decay constant (ln(2) / half-life)
t = the time elapsed (1.719 x 10^4 years)
First, let's find the decay constant (k) using the half-life:
k = ln(2) / half-life
k = ln(2) / 5730
Now we can substitute the values into the formula and calculate the remaining amount:
N = 144 * e^(-k * t)
N = 144 * e^(-ln(2) / 5730 * 1.719 x 10^4)
Calculating this expression gives us:
N ≈ 93.81 g
Therefore, approximately 93.81 grams of the carbon-14 sample will remain after 1.719 x 10^4 years.
To determine the amount of carbon-14 that will remain after a certain time, we can use the formula for exponential decay:
N(t) = N(0) * (1/2)^(t / T)
Where:
N(t) is the amount remaining after time t
N(0) is the initial amount
t is the time that has passed
T is the half-life of the substance
In this case, we have N(0) = 144 g, t = 1.719 × 10^4 years, and T = 5730 years.
Substituting these values into the formula, we can calculate the amount of carbon-14 remaining:
N(t) = 144 * (1/2)^(1.719 × 10^4 / 5730)
To simplify the calculation, divide the exponent by the half-life:
N(t) = 144 * (1/2)^(1.719 × 10^4 / 5730) = 144 * (1/2)^(3)
Now we need to calculate (1/2)^3, which is equal to 1/2 * 1/2 * 1/2 = 1/8.
N(t) = 144 * (1/8) = 18 g
Therefore, after 1.719 × 10^4 years, only 18 grams of the 144-gram sample of carbon-14 will remain.