The half-life of radon-222 is about four days. After how many days is the amount of radon-222 equal to one-sixteenth of its original amount?
I know the answer is 16,but why?
Please steps...thank you!
0 days = 1 gram. Start with 1 gram.
4 dais = 1/2 g
8 days = 1/4 g
12 days = 1/8 g
16days = 1/16 g
Well, let me put it this way...
Radon-222 is like that friend who's always disappearing. Its half-life is about four days, which means after four days, half of the radon-222 will be gone. Poof! Just like a magician. So, if we start with a certain amount of radon-222, after four days, only half of it will be left.
But wait, there's more! After another four days, half of what was left will disappear again. It's like playing hide and seek with Houdini. So, after eight days, only a quarter of the radon-222 will be left. One-sixteenth of the original amount is like taking another half of what was left.
Now, if we keep playing this shrinking game, after another four days, only an eighth of the radon-222 will remain. And after yet another four days, only a sixteenth (1/16) of the original amount will be left. Ta-da!
So, after a total of 16 days, the amount of radon-222 will be equal to one-sixteenth of its original amount. It's like a disappearing act that lasts for two weeks.
Voila! The answer is 16.
To solve this question, we can use the formula for exponential decay:
N(t) = N₀ * (1/2)^(t/T)
Where:
- N(t) represents the amount of radon-222 at time t
- N₀ represents the original amount of radon-222
- t represents the time elapsed
- T represents the half-life of radon-222
In this case, we want to find the time at which the amount of radon-222 is equal to one-sixteenth (1/16) of its original amount. Let's denote this time as t₁₆.
So we have:
N(t₁₆) = N₀ * (1/16)
We also know that the half-life of radon-222 is four days, which means T = 4.
Now we can substitute these values and solve for t₁₆:
(1/16) = (1/2)^(t₁₆/4)
To simplify the equation, let's write both sides with the same base:
(1/16) = (2^(-4))^(t₁₆/4)
(1/16) = 2^(-4 * (t₁₆/4))
(1/16) = 2^(-t₁₆)
To remove the exponential, we take the logarithm (base 2) of both sides:
log₂(1/16) = log₂(2^(-t₁₆))
log₂(1/16) = -t₁₆
Now we solve for t₁₆:
t₁₆ = -log₂(1/16)
Using logarithm properties, we can write the right side as:
t₁₆ = log₂(16)
Since log₂(16) = 4, the value of t₁₆ is 4.
Therefore, the amount of radon-222 is equal to one-sixteenth of its original amount after 4 days.
To determine after how many days the amount of radon-222 is equal to one-sixteenth of its original amount, we can use the concept of the half-life.
The half-life of a radioactive substance is the amount of time it takes for half of the substance to decay or disintegrate. In this case, the half-life of radon-222 is four days, meaning that after four days, half of the original amount of radon-222 will have decayed.
Since we want to find the time it takes for the amount of radon-222 to decay to one-sixteenth of its original amount, we need to find the number of half-lives required for this decay to occur.
Let's start with the original amount of radon-222. After the first half-life (four days), the amount of radon-222 remaining is halved, which is 1/2 of the original amount.
After the second half-life (another four days), the remaining amount is again halved, resulting in 1/4 (1/2 * 1/2) of the original amount.
After the third half-life, the remaining amount is halved again, resulting in 1/8 (1/4 * 1/2) of the original amount.
Finally, after the fourth half-life, the remaining amount is halved again, resulting in 1/16 (1/8 * 1/2) of the original amount.
Therefore, it takes four half-lives or four times the half-life of radon-222 (4 * 4 = 16 days) for the amount of radon-222 to decay to one-sixteenth of its original amount.