If the half-life of uranium-235 is 7.04 (times) 10 (to the eight power) 8 y and 12.5 g of uranium-235 remain after 2.82 (times) 10 (to the ninth power) y, how much of the radioactive isotope was in the original sample?
You need the rate constant which is
k = 0.694/t1/2>/sub>
Substitute k into the equation below.
ln(No/N) = kt.
No is what you wnat to know.
N = 12.5g from the problem.
k from above.
t = 2.82E9 yr.
Substitute and solve for N.
To find out how much of the radioactive isotope was in the original sample, we can use the concept of half-life.
The half-life of uranium-235 is given as 7.04 × 10^8 years. This means that after every 7.04 × 10^8 years, half of the original sample will decay.
We are given that after 2.82 × 10^9 years, 12.5 g of uranium-235 remain. We need to figure out how much uranium-235 was in the original sample.
Let's start by determining how many half-lives have passed. We can divide the total time that has passed (2.82 × 10^9 years) by the half-life of uranium-235 (7.04 × 10^8 years):
Number of half-lives = (2.82 × 10^9 years) / (7.04 × 10^8 years)
= 4
This means that 4 half-lives have passed.
Since each half-life reduces the amount by half, we can express the remaining amount of uranium-235 in terms of the original sample size (let's call it X):
12.5 g = X * (1/2)^4
12.5 g = X * 1/16
To isolate X, we can multiply both sides of the equation by 16:
16 * 12.5 g = X
200 g = X
Therefore, the original sample of uranium-235 contained 200 grams.