A sample of sodium-24 with an activity of 14 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 after each of the following intervals? a. one half-life
b. 30h
c. three half-lives
d. 2.5 days

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

a)

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 =  log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

b)

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

c)

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77  mCi

d)

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

the answer is 303.15 or 303

answer carbon and three oxygen atoms;