If 200. mg of fluorine-18 is shipped at 7:00 A.M., how many milligrams of the radioisotope are still active when the sample arrives at the radiology laboratory at 4:10 P.M.?
Well, if you manage to keep the radioactive fluorine-18 still for that long, you should probably apply for a job in the circus! But to answer your question, the half-life of fluorine-18 is about 110 minutes. So, let's calculate the approximate amount of fluorine-18 that would still be active when it arrives at 4:10 P.M. Hmm... carrying the 1, multiplying by the square root of the mass-to-energy ratio... drumroll... I estimate about 0.0076328 mg of fluorine-18 would still be active. But don't worry, it won't bite!
To determine the amount of fluorine-18 that remains active when the sample reaches the radiology laboratory, we need to consider the half-life of fluorine-18 and the time elapsed between the shipment and arrival.
The half-life of fluorine-18 is approximately 109.7 minutes.
Step 1: Calculate the number of half-lives that have passed
Since the shipment was made at 7:00 A.M. and the arrival time is 4:10 P.M., the time elapsed is 9 hours and 10 minutes.
Step 2: Convert the elapsed time to minutes
9 hours x 60 minutes = 540 minutes (from hours to minutes)
540 minutes + 10 minutes = 550 minutes (including the additional 10 minutes)
Step 3: Divide the elapsed time by the half-life
550 minutes / 109.7 minutes = 5.012 half-lives (rounded to three decimal places)
Step 4: Calculate the remaining activity based on the number of half-lives
The remaining activity is calculated using the formula: final activity = initial activity × (1/2)^n, where n is the number of half-lives that have passed.
final activity = 200 mg × (1/2)^5.012
Step 5: Calculate the result
final activity ≈ 200 mg × 0.03125
final activity ≈ 6.25 mg (rounded to two decimal places)
Therefore, when the sample arrives at the radiology laboratory at 4:10 P.M., approximately 6.25 milligrams of fluorine-18 are still active.
To determine the amount of fluorine-18 that is still active when it arrives at the radiology laboratory, we need to consider its half-life.
The half-life of fluorine-18 is approximately 110 minutes. This means that every 110 minutes, half of the radioisotope decays and becomes inactive.
To find out how many half-lives have passed from 7:00 A.M. to 4:10 P.M., we need to calculate the elapsed time.
From 7:00 A.M. to 4:00 P.M., a total of 9 hours have passed. Since each hour has 60 minutes, this amounts to 9 x 60 = 540 minutes.
From 4:00 P.M. to 4:10 P.M., an additional 10 minutes have passed.
Therefore, in total, it took 540 + 10 = 550 minutes for the sample to arrive at the radiology laboratory.
Now, we divide this total time by the half-life of fluorine-18 (110 minutes) to determine the number of half-lives that have passed:
550 minutes / 110 minutes = 5 half-lives.
Each half-life halves the amount of the radioisotope, so after 5 half-lives, the remaining activity is given by:
Remaining activity = Initial activity × (1/2)^(number of half-lives).
The initial activity is 200 mg, so we can calculate the remaining activity as follows:
Remaining activity = 200 mg × (1/2)^5 = 200 mg × (1/2)^5 = 200 mg × (1/32) = 6.25 mg.
Therefore, when the sample arrives at the radiology laboratory at 4:10 P.M., there are approximately 6.25 mg of fluorine-18 still active.
half life = 109.8minutes
minutes from 7am to 4:10= 9*60+10
Amountleft= 200mg*e^(-.693*minutes/109.8)