Answer:
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.
