After 150 thousand years, only 1/8 of the original amount of a particular radioactive waste will remain. The half-life of this radioactive waste is how many thousand years?
N = (No)e^(-kt)
where
No is the original amount
k is constant
t is time
substitute,
(1/8)*No = No * e^(-k*150000)
the No cancels:
1/8 = e^(-150000k)
ln (1/8) = -150000k
k = -150000/ ln(1/8)
after solving for k, substitute to this equation and get the value of t (half-life):
1/2 * No = No * e^(-kt)
What does the math symbols such as ^ and * mean? I still don't understand exactly what your trying to solve for. Am I trying to figure out the half-life of radioactive waste? Am I trying to figure out how many thousand years it takes in a half-life of radioactive waste? What does the term in math mean when saying "half-life"? Is the 1/8 suppose to stay as a fraction or be turned into a decimal or percent?
the symbol ^ means "raise to". the symbol * means "multiplied by" as in for example
3^2 = 3*3 = 9
Yes, we're looking for the half-life. Half-life is the time it takes for a substance to achieve 1/2 of its original amount when it undergoes radioactive decay. the formula used for radioactive decay is
N = (No)*e^(-k*t)
where
N = remaining amount after time, t
No = the original amount
k = some constant
t = time
since we don't know the k value (nothing is given), as well as the original amount (the No), express N in terms of No so we can cancel them on both sides. that's why the equation becomes
(1/8)*No = No * e^(-k*150000)
It is stated in the problem that after 150000 years (this is the variable t), the remaining amount will only be 1/8 of the original (the original is No, right?).
since the equation above contains No term, we can cancel them:
(1/8) = e^(-k*150000)
then get the natural log of both sides. That will cancel the e on the right side, leaving only the exponent:
ln (1/8) = ln (e^(-k*150000))
ln (1/8) = -150000k
k = -150000/ ln(1/8)
Get a calculator and solve the above equation to get k.
Now, after getting k, you need to substitute it to the original equation:
N = (No)*e^(-k*t)
The N here will be equal to (1/2)*No since half-life (remember the definition?). Again, No will be cancelled on both sides. The equation becomes:
1/2 * No = No * e^(-kt)
1/2 = e^(-k*t)
ln (1/2) = ln(e^(-k*t))
ln (1/2) = -k*t
k = (-ln(1/2)) / t
now you know the value of k, solve for t.
i mean,
t = (-ln(1/2)) / k
we're solving for t. the units are in years.
Wow - that's a lot of work.
each half-life diminishes the amount by half.
1/8 = 1/2 * 1/2 * 1/2
So, 150,000 years is 3 half-lives
The half-life is thus 50,000 years.
To find the half-life of a radioactive substance, we need to use the decay formula:
N = N₀ * (1/2)^(t / t₁/₂)
Where:
N is the remaining amount of the substance after time t
N₀ is the initial amount of the substance
t is the time that has passed
t₁/₂ is the half-life of the substance
In this case, we know that after 150 thousand years, only 1/8 (or 1/2^3) of the original amount will remain. So, we can write the equation as:
1/8 = 1 * (1/2)^(150 / t₁/₂)
To find the half-life (t₁/₂), we need to solve this equation.
First, let's simplify the equation:
1/8 = (1/2)^(150 / t₁/₂)
Now, let's take the logarithm of both sides to solve for the exponent:
log(1/8) = log((1/2)^(150 / t₁/₂))
Using the logarithmic property, we can move the exponent down as a coefficient:
log(1/8) = (150 / t₁/₂) * log(1/2)
The logarithm of 1/8 (or 2^(-3)) is -3:
-3 = (150 / t₁/₂) * log(1/2)
We want to isolate t₁/₂, so let's divide both sides by log(1/2):
-3 / log(1/2) = 150 / t₁/₂
Now, we can solve for t₁/₂ by isolating it:
t₁/₂ = (150 * log(1/2)) / -3
Calculating this expression will give us the half-life of the radioactive waste in thousand years.