Question

Suppose the half- life of a certain radioactive substance is 12 days and there are 18 grams present initially. Find the time when there will be 20% of the initial substance remaining.

Answer 1

Answer:

4.8 days i hope it will help you

Step-by-step explanation:

18 g is active for 24 days

then for 1 g it will active for 24/18

and for 20 %

20% of 18 g = 3.6

the time is equal to 24*3.6/18

that is equal to 4.8 days

Answer 2

Answer:

t ≈ 27.86 days

Step-by-step explanation:

$a^{b}=c$ ⇔ $log_{a}c = b$

~~~~~~~

t ≈ 27.86 days