Assume a substance has a half-life of 11 years and the initial
amount is 144 grams.
How much remains at the end of 5 years?
How long will it be until only 30 % remains?
thanks
1.028
To determine how much of the substance remains at the end of 5 years, we need to use the formula for exponential decay:
A = A₀ * (0.5)^(t / T)
Where:
- A is the amount remaining after time t
- A₀ is the initial amount
- t is the given time elapsed
- T is the half-life of the substance
Let's calculate it:
1. Calculate the amount remaining after 5 years:
A = 144 * (0.5)^(5 / 11)
A ≈ 82.60 grams
So, approximately 82.60 grams of the substance will remain at the end of 5 years.
2. To find out how long it will take until only 30% remains, we can re-arrange the formula:
0.3 = A₀ * (0.5)^(t / T)
Let's solve for t:
0.3 / A₀ = (0.5)^(t / T)
Take the logarithm base 0.5 of both sides:
log₁₀(0.3 / A₀) = (t / T) * log₁₀(0.5)
Divide both sides by log₁₀(0.5):
(t / T) = log₁₀(0.3 / A₀) / log₁₀(0.5)
Now, substitute the given values:
(t / 11) = log₁₀(0.3 / 144) / log₁₀(0.5)
Multiply both sides by 11:
t ≈ 11 * (log₁₀(0.3 / 144) / log₁₀(0.5))
Using a calculator, we find:
t ≈ 23.14 years
Therefore, it will take approximately 23.14 years until only 30% of the substance remains.
If A(t)=quantity in grams left after t years,
A0 is the initial quantity in grams, then
R=rate of decay (0<R<1)
A(t)=A0 R^t
For a half life of 11 years,
t=11
A(t)=0.5A0
R^t=A(t)/A0=0.5
Take ln and put t=11
11ln(R)=0.5
Solve for R.
You can solve the second part similar to the first. Post your answers for checking if you wish.