An isotope has a 12.4 hour half life. If it has an activity of 30mCi at noon today, what was its activity at 3 p.m. yesterday?
the time involved is 9+12 = 21 hours, or 1.69 half-lives
so, we need to find P where
30 = P*(1/2)^1.69
P = 96.8
Well, let me calculate that for you, but first, let me ask you a question. Why did the isotope go to therapy? It had a half life crisis! Alright, let's get down to business.
To determine the activity at 3 p.m. yesterday, we need to calculate the number of half-lives that have passed in that time frame. From 3 p.m. yesterday to noon today, 21 hours have passed.
Now, since the half-life is 12.4 hours, we can divide the time interval by the half-life to find the number of half-lives. In this case, 21 hours divided by 12.4 hours gives us approximately 1.69 half-lives.
Now, let's apply some mathematical magic. Since each half-life reduces the activity by half, we'll take the initial activity of 30mCi and divide it by 2 raised to the power of 1.69.
By crunching the numbers, we find that the activity at 3 p.m. yesterday was approximately 30mCi divided by about 2.51. So, drumroll please... the activity at 3 p.m. yesterday was approximately 11.95mCi.
But hey, don't sweat it if the math gets a bit complicated. Just remember that the isotope had a half-life crisis and needed some professional help to figure things out!
To calculate the activity at a specific time, we need to determine the number of half-lives that have elapsed between two given times. Then, we can use the half-life equation to find the activity at the second time.
Let's break down the problem step by step:
Step 1: Calculate the number of half-lives elapsed between the two given times.
Since the isotope has a half-life of 12.4 hours, we need to determine how many 12.4-hour intervals there are between 3 p.m. yesterday and noon today.
Number of hours = 24 (hours in a day) + 12 (hours from noon today to 3 p.m. yesterday) = 36 hours
Number of half-lives = 36 hours / 12.4 hours per half-life = 2.9032 or approximately 3 half-lives.
Step 2: Determine the remaining fraction of the isotope.
The remaining fraction can be calculated using the formula:
Remaining fraction = (1/2)^(number of half-lives)
Remaining fraction = (1/2)^3 = 1/8
Step 3: Calculate the activity at 3 p.m. yesterday.
Given that the activity at noon today is 30mCi, we can use the remaining fraction to calculate the activity at 3 p.m. yesterday:
Activity at 3 p.m. yesterday = Remaining fraction * Activity at noon today
Activity at 3 p.m. yesterday = (1/8) * 30mCi = 3.75mCi
Therefore, the activity at 3 p.m. yesterday was approximately 3.75mCi.
To find the activity of the isotope at 3 p.m. yesterday, we can use the concept of half-life and apply it to the given information.
First, let's determine the number of half-lives between yesterday at 3 p.m. and today at noon. We know that the half-life of the isotope is 12.4 hours. Therefore, the time between yesterday at 3 p.m. and today at noon is 21 hours.
To calculate the number of half-lives, we divide the total time by the half-life:
Number of half-lives = Total time / Half-life
Number of half-lives = 21 hours / 12.4 hours
Approximately, the number of half-lives is 1.69 (rounded to two decimal places). Since you can't have a fraction of a half-life, we will consider it as one whole half-life.
Now, since we have one half-life, we know that the activity is reduced by half after each half-life. Therefore, the activity at 3 p.m. yesterday would be half of the current activity at noon today.
Given that the activity at noon today is 30 mCi, we can calculate the activity at 3 p.m. yesterday:
Activity at 3 p.m. yesterday = Activity at noon today / 2
Activity at 3 p.m. yesterday = 30 mCi / 2
Activity at 3 p.m. yesterday = 15 mCi
So, the activity of the isotope at 3 p.m. yesterday would be approximately 15 mCi.