a sample of sodium-24 with an activity of 12 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24 has a half-life of 15 h, what is the activity of the sodium after 2.5 days. please show work.

1). half life = ln(2) / lambda

where lambda is the radioactive decay constant

2). A(t) = 12mCi * e ^ - (lambda * t)

where t is time, A(t) is activity as a function of time

15 hr = ln2 / lambda

solve for lambda

convert 2.5 days to hours
Then t = 2.5 days * (24 hrs / day)

Plug these numbers into A(t)

To find the activity of sodium-24 after 2.5 days, we need to calculate how much of the sample has decayed during that time.

First, let's convert 2.5 days into hours. Since there are 24 hours in a day, we multiply 2.5 by 24:

2.5 days * 24 hours/day = 60 hours

Next, we can use the radioactive decay formula to calculate the fraction remaining:

N(t) = N(0) * (1/2)^(t / T)

Where:
N(t) = remaining activity after time t
N(0) = initial activity
t = time elapsed
T = half-life

Let's plug in the given values:
N(0) = 12 mCi (initial activity)
t = 60 hours (time elapsed)
T = 15 hours (half-life)

N(60) = 12 mCi * (1/2)^(60/15)

Now, we simplify the equation:

N(60) = 12 mCi * (1/2)^4

N(60) = 12 mCi * 1/16

N(60) = 0.75 mCi

Therefore, the activity of the sodium-24 after 2.5 days (60 hours) is 0.75 mCi.