Answer:
B
Step-by-step explanation:
The decay of U-235 follows an exponential decay model, which can be described by the equation:
N = N0 * (1/2)^(t/T)
where N is the amount of U-235 at a given time t, N0 is the initial amount of U-235, t is the elapsed time, and T is the half-life of U-235.
In this case, we know that the sample has undergone two half-lives, which means that the elapsed time is:
t = 2 * T
We also know that the amount of U-235 in the sample is 0.15 ug. We can use this information to solve for the initial amount of U-235 (N0):
N = N0 * (1/2)^(t/T)
0.15 = N0 * (1/2)^(2)
0.15 = N0 * (1/4)
N0 = 0.15 / (1/4)
N0 = 0.60 ug
Therefore, the amount of U-235 that was originally present in the sample before it decayed was 0.60 ug. The answer is B) 0.60 ug.
Even more:
It's not A) 0.30 ug because if the sample originally contained 0.30 ug of U-235 and underwent two half-lives, the amount of U-235 remaining would be:N = N0 * (1/2)^(t/T)
N = 0.30 * (1/2)^(2)
N = 0.30 * (1/4)
N = 0.075 ug
This means that the amount of U-235 in the sample after two half-lives would be 0.075 ug, which is not consistent with the given information that the sample contains 0.15 ug of U-235.
Therefore, the correct answer is B) 0.60 ug, which is the initial amount of U-235 that would have been present in the sample before it decayed.