A radioactive isotope(h=4.5days) was prepared. This isotope was used 14 days later when it had an activity of 6000000 Bq. What was the activity of this isotope when initially prepared?
(Initial activity)*(0.5)^(14/4.5) = 6*10^7 Bq
Inital activity = 6*10^7*2^3.111
= 5.2*10^8 Bq
The answer on my handout is 5.18 x 10^7
I read your 6000000 incorrectly as 6*10^7. Using commas or scientific notation would have helped me read the string of zeros.
The handout answer is correct.
To find the initial activity of the radioactive isotope, we can use the concept of half-life.
The half-life of a radioactive isotope is the time it takes for the activity to decrease by half. In this case, the half-life of the isotope is given as 4.5 days.
To calculate the initial activity, we can use the formula:
A = A0 * (1/2)^(t / h)
Where:
- A is the final activity (6000000 Bq)
- A0 is the initial activity (which we want to find)
- t is the time elapsed (14 days)
- h is the half-life (4.5 days)
Now, let's plug in the values and solve for A0:
6000000 = A0 * (1/2)^(14 / 4.5)
To isolate A0, we can divide both sides of the equation by (1/2)^(14 / 4.5):
6000000 / (1/2)^(14 / 4.5) = A0
Simplifying further:
6000000 / (1/2)^(14 / 4.5) ≈ A0
Using a calculator, we can evaluate the right-hand side of the equation:
A0 ≈ 6000000 / (1/2)^(14 / 4.5)
After evaluating the expression, we find that the initial activity of the isotope was approximately 47,833,790 Bq.