An isotope of lead, 201Pb, has a half-life of 8.4 hours. How many hours ago was there 40% more of the substance? (Round your answer to one decimal place.)
____hr
If there was 40% more, then that's 1.4 times what's there now.
So, starting with an original amount of 1, there is now 1/1.4 of it left.
(1/2)^(t/8.4) = 1/1.4
Now just solve for t hours.
Note: It will be less than 8.4, since 8.4 hours ago, there was 100% more.
not sure how to find the t, tried using the log approach but can't get the right answer.
8.4
To determine how many hours ago there was 40% more of the substance, we need to first find the amount of the substance at that time.
Let's assume the initial amount of the substance was 100 units.
After one half-life (8.4 hours), the amount of the substance remaining would be 50 units, because half of the substance decays every half-life.
If we want to find the time when there was 40% more of the substance, we need to find 40% of the initial amount (100 units) and add it to the initial amount.
40% of 100 units = 0.4 * 100 = 40 units
Adding this to the initial amount, we get 100 units + 40 units = 140 units.
Now, we need to find out how many half-lives it takes for the substance to decay to 140 units.
We can set up the equation:
50 units * (1/2)^(n) = 140 units
where n is the number of half-lives.
Dividing both sides of the equation by 50 units and taking logarithms, we can solve for n.
(1/2)^(n) = 140/50
n * ln(1/2) = ln(140/50)
n = ln(140/50) / ln(1/2)
Using a calculator, we find that n is approximately 1.129.
Since each half-life is 8.4 hours, we multiply n by 8.4 to find how many hours ago there was 40% more of the substance.
1.129 * 8.4 = 9.49 hours
Therefore, approximately 9.5 hours ago there was 40% more of the substance.