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((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.