In 90 seconds, the mass of a radioactive element reduces to \(\frac{1}{16}\) of its original value. Determine the half- life of the element.
15.0s
18.0s
22.5s
30.0s
45.0s
Correct answer is C
\(\frac{1}{16} = (\frac{1}{2})^n\)
\((\frac{1}{2})^4 = (\frac{1}{2})^n\)
\(n = 4\)
90 secs = 4 half lives
1 half life = \(\frac{90}{4}\) secs
= 22.5 secs