Question

In: Chemistry

Technetium-99 is used as a radioisotope in a variety of medical applications . if a patient...

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.

Solutions

Expert Solution

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.


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