The initial activity of an iridium-192 sample was 3000 dpm and dropped to 375 dpm after 222 days.What is the half-life of this radionuclide?

Question

Quizplus · Accepted Answer

Using the formula for radioactive decay:
N=N0(1/2)^(t/T)
where N0 is the initial activity, N is the activity after time t, and T is the half-life.
Plugging in the values given:
375=3000(1/2)^(222/T)
Dividing both sides by 3000:
0.125=(1/2)^(222/T)
Taking the natural logarithm of both sides:
ln(0.125)=(222/T)ln(1/2)
Simplifying:
ln(0.125)=-0.693(T/222)
Solving for T:
T=-222(ln(0.125)/0.693)=74.0 days.

Therefore, the correct answer is B) 74.0 days.

The Initial Activity of an Iridium-192 Sample Was 3000 Dpm