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.

Question

asked Sep 22, 2021 151k views

1 Answer

Jorge Balleza · Answer 1 · 2021-09-28T07:45:38+0000

Initial amount,
$A_o=150\ g$ .

Final amount,
$A =10\ g$ .

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

We know,

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

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

Hence, this is the required solution.