If 18g of a radioactive substance are present initially and 8 yr later only 9 g remain, how much of the substance will be present after 20 yr?
Well, when it comes to radioactive substances, they sure do know how to disappear over time! So, if we start with 18g and after 8 years only 9g remain, we can tell that half of the substance has vanished into thin air.
Now, assuming this radioactive vanishing act continues at the same rate, after 20 years we can expect another half of what's left to vanish. So, if we currently have 9g left, we can estimate that in 20 years, half of that will be gone too.
If only radioactive substances could pull off a magic trick to leave us with zero grams! But using our radioactive mathematics, we can calculate that after 20 years, approximately 4.5g will remain.
Keep in mind though, this is a simplified estimate that assumes the rate of decay remains constant over time. So, don't be startled if reality throws in a curveball or two. Stay prepared, wizard of radioactive math!
To find out how much of the substance will be present after 20 years, we can use the concept of radioactive decay. The rate of decay of a radioactive substance is often measured using a parameter called the half-life. The half-life represents the time it takes for half of the substance to decay.
In this case, we know that 18 grams of the substance are initially present and after 8 years, only 9 grams remain. This implies that the substance has a half-life of 8 years, as half of the initial amount has decayed.
To determine the rate of decay, we can calculate the decay constant (k) using the half-life formula:
k = (ln(2)) / t₁/₂,
where k is the decay constant and t₁/₂ is the half-life. In this case, t₁/₂ = 8 years.
Substituting these values into the formula, we can calculate k:
k = (ln(2)) / 8 ≈ 0.08664
Now, we can use the decay equation to find the amount of the substance remaining after 20 years:
A = A₀ * e^(-kt),
where A is the final amount, A₀ is the initial amount, k is the decay constant, and t is the time.
Substituting the given values into the equation:
A = 18 * e^(-0.08664 * 20)
Using basic algebra and a calculator, we find:
A ≈ 5.654 grams
Therefore, after 20 years, approximately 5.654 grams of the radioactive substance will remain.
half gone in 8 years?
x = Xi e^-kt
.5 = e^-8k
ln .5 = -.693 = - 8 k
so
k = .08664
x = 18 e^-.08664 t
when t = 20
x = 3.18 grams
clearly the half-life is 8 years. So, after t years, the remaining fraction is
18*2^(-t/8)
So, plug in t=20