Question

Iodine-125 is radioactive and has a half life of 60.25 days. How much of a 7.40 mg sample would be left after 243. days? Round your answer to 2 significant digits. Also, be sure your answer has a unit symbol. 0 0.0 X ?

Answer 1

Answer:

Initial mass of Iodine-125 = 7.40 mg

Half life of iodine-125 = t_1/2 = 60.25 days

Radio active decay follows first order kientics.

Integrated rate law of first order kientics is given by :

$N=N_o\times e^{-kt}\\\\ k=\frac{0.693}{t_{\frac{1}{2}}}$

where:
No = Initial mass of an isotope
N = Mass of an isotope left after the time t
t_1/2 = Half-life of an isotope
k = Decay constant

According to question we have:

No = 7.40 mg
N = ?

t = 243 days
t_1/2 = 60.25 days

Now put all the given values in this formula, we get

N = 7.40mg xe-60.25days x 243days)

On solving for N, we get N:

$N = 0.452 mg \approx 0.45 mg$

0.45 mg of iodine-125 will be left after 243 days.