In: Chemistry
Technetium-99 is used as a radioisotope in a variety of medical applications . if a patient receives a dosage of 6.00 mg at 8 am on thursday , then how much would remain at 4pm on friday afternoon? the half life of Tc-99 is 6.0 hours.
Radio active decay is a first order reaction.
For first order reaction,
half life t1/2 = 0.693 /k where k is rate constant
k = 0.693/ t1/2 --- Eq (1)
k = 1/t ln{ [A]o/[A]t} -----Eq (2)
From Eqs (1) and (2),
0.693/ t1/2 = (1/t) ln {[A]o/ [A]t} ------Eq (3)
Given that
half life of Tc-99 t1/2 = 6.0 hrs
time t = 32 hrs [ 8am to next day4 pm = 32 hrs ]
Initial amount of Tc-99 = 6.0 mg
Final amount of Tc-99 [A]t = ?
Substitute all the values in Eq (3),
0.693/ t1/2 = (1/t) ln {[A]o/ [A]t}
[(0.693)/(6 hrs)] = (1/32 hrs) ln {6 mg/ [A]t}
ln {6 mg/ [A]t} = [0.693/6]x 32
- ln {[A]t/ 6 mg } = [0.693/6]x 32
ln {[A]t/ 6 mg } = - [0.693/6]x 32
[A]t/ 6 mg = e - [0.693/6]x 32
[A]t = (6 mg) .e - [0.693/6]x 32
= 0.148 mg
[A]t = 0.148 mg
Final amount of Tc-99 = 0.148 mg
Therefore,
0.148 mg of Tc-99 would would remain at 4pm on friday afternoon.