how many half-lives have passed if 87.5% of a substance has decomposed? how many if 99.999% has decomposed?
Well, well, well, let me think about that, or at least attempt to. If 87.5% of a substance has decomposed, that means there's only 12.5% left. Since each half-life reduces the amount by half, it would take two half-lives for it to go from 100% to 25%, and another two half-lives for it to go from 25% to 6.25%, leaving us with four half-lives in total.
Now, let's tackle the second one. If a whopping 99.999% has decomposed, that means there's only 0.001% remaining. In order to go from 100% to 0.001%, we would need to go through (drumroll, please) about 13 half-lives!
So, to sum it up, if 87.5% has decomposed, we have four half-lives under our belts, and if 99.999% has decomposed, we'd be looking at about thirteen hilarious half-lives.
To find the number of half-lives that have passed, we can use the formula:
Number of half-lives = (log(initial amount / final amount)) / (log(2))
For the first scenario where 87.5% of the substance has decomposed, the remaining amount after decomposition is 12.5% (100% - 87.5%). Therefore, the initial amount is 100% and the final amount is 12.5%.
Number of half-lives = (log(100% / 12.5%)) / (log(2))
Number of half-lives = (log(8)) / (log(2))
Number of half-lives ≈ 3
Therefore, if 87.5% of the substance has decomposed, approximately 3 half-lives have passed.
For the second scenario where 99.999% of the substance has decomposed, the remaining amount is 0.001% (100% - 99.999%). Therefore, the initial amount is 100% and the final amount is 0.001%.
Number of half-lives = (log(100% / 0.001%)) / (log(2))
Number of half-lives = (log(100000)) / (log(2))
Number of half-lives ≈ 16.61
Therefore, if 99.999% of the substance has decomposed, approximately 16.61 half-lives have passed.
To determine the number of half-lives that have passed, we need to use the formula for exponential decay:
Decayed amount = Initial amount × (1/2)^(number of half-lives)
Let's solve the problem step by step:
1. If 87.5% of a substance has decomposed, it means that 12.5% (100% - 87.5%) remains unchanged. Therefore, we are looking for the number of half-lives needed for 12.5% of the substance to remain.
2. We can set up the following equation:
0.125 = 1 × (1/2)^(number of half-lives)
3. To isolate the exponential term, we divide both sides of the equation by 1:
0.125 / 1 = (1/2)^(number of half-lives)
4. Simplifying the left side gives us:
0.125 = (1/2)^(number of half-lives)
5. To solve for the number of half-lives, we take the logarithm base 2 of both sides:
log2(0.125) = log2((1/2)^(number of half-lives))
6. Using the logarithmic property log_b(x^y) = y · log_b(x), the equation becomes:
log2(0.125) = number of half-lives · log2(1/2)
7. Since log2(1/2) equals -1, we can simplify further:
log2(0.125) = number of half-lives × (-1)
8. Rearranging the equation gives us:
number of half-lives = log2(0.125) / (-1)
9. Evaluating the expression using a calculator, we find:
number of half-lives = -3
Therefore, 3 half-lives have passed if 87.5% of the substance has decomposed.
Now let's move on to the second part of the question:
1. If 99.999% of a substance has decomposed, it means that 0.001% (100% - 99.999%) remains unchanged. We need to find the number of half-lives required for 0.001% to remain.
2. Setting up the equation:
0.00001 = 1 × (1/2)^(number of half-lives)
3. Dividing by 1 to isolate the exponential term:
0.00001 / 1 = (1/2)^(number of half-lives)
4. Simplifying the left side:
0.00001 = (1/2)^(number of half-lives)
5. Taking the logarithm base 2 of both sides:
log2(0.00001) = log2((1/2)^(number of half-lives))
6. Applying the logarithmic property:
log2(0.00001) = number of half-lives · log2(1/2)
7. Since log2(1/2) equals -1:
log2(0.00001) = number of half-lives × (-1)
8. Rearranging the equation:
number of half-lives = log2(0.00001) / (-1)
9. Calculating the expression using a calculator:
number of half-lives ≈ 16.6
Therefore, approximately 16.6 half-lives have passed if 99.999% of the substance has decomposed.
Let x be the number of half-lives.
For 87.5%, 12.5% = 1/8 is left
2^-x = 1/8
x = 3
For 99.999%, 10^-5 is left
2^-x = 10^-5
x = 16.6