Question

If 150 grams of a radioactive isotope are present at 2:00 PM and 10 grams remain at 6:00 PM (the same day), what is the half-life of the isotope? Round to two decimal places.

Answer 1

Initial amount, $A_o=150\ g$ .

Final amount, $A =10\ g$ .

Time taken, t = 6:00 - 2:00 = 4 hour.

We know,

$A=A_o(\dfrac{1}{2})^{\dfrac{t}{h}}\\\\10 = 150 \times \dfrac{1}{2}^{\dfrac{4}{h}}\\\\2^{\dfrac{4}{h}}=15\\\\\dfrac{4}{h}= log_215\\\\h = \dfrac{4}{log_215}\\\\h = 1.024 \ hours$

Therefore, the half-life of the isotope is 1.024 hours.

Hence, this is the required solution.