How much of a radioactive material with a half-life of 12,000 years is left after 36,000 years of decay?
That is three half-lives?
first half life 1/2 remaining
second half life 1/4 remaining
third half life 1/8 remaining
amountrmaing=0riginalamount (1/2)^(t/th)
where t is time, th is halflife time.
I know I'm repeating myself, Angie; however, all of these problems can be worked with the formula I gave you and solved an example problem for you a couple of days ago. You need to learn to do these yourself instead of posting and obtaining answers from a third party. DrWLS, Bob Pursley, and I won't be with you when it comes test time.
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To find out how much of a radioactive material remains after a certain amount of time, you can use the exponential decay equation:
N(t) = N₀ * (1/2)^(t / T)
Where:
- N(t) is the amount of material remaining after time t
- N₀ is the initial amount of material
- t is the elapsed time
- T is the half-life of the radioactive material
In this case, we have a half-life of 12,000 years and a time of 36,000 years. We need to find the amount of material left after 36,000 years, so we can plug these values into the equation:
N(t) = N₀ * (1/2)^(t / T)
N(36,000) = N₀ * (1/2)^(36,000 / 12,000)
Let's calculate it step-by-step:
1. Convert the time to half-lives: 36,000 years / 12,000 years = 3 half-lives.
2. Plug this into the equation: (1/2)^(3).
Evaluating this expression, we find:
(1/2)^3 = 1/2 * 1/2 * 1/2 = 1/8.
So, the equation becomes:
N(36,000) = N₀ * (1/8)
To determine the remaining amount, we need to know the initial amount of the radioactive material. Once we have this value, we can multiply it by 1/8 to find the amount remaining after 36,000 years.