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See photo please!! I only have 10 mins

See photo please!! I only have 10 mins-example-1
User TimT
by
7.9k points

2 Answers

6 votes

Answer:

0.30 ug of U-235

Step-by-step explanation:

N = N0 * (1/2)^(1/2)

0.15 ug = N0 * (1/2)^(1/2)

N0 = 0.15 ug / (1/2)^(1/2)

N0 = 0.30 ug

User Shreyas D
by
8.4k points
4 votes

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.

User Jeff Noel
by
8.3k points

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