A radioactive substance has a half-life of 20 hours. What fraction of the original radioactive nuclide will remain after 80 hours?
- 1/32
-
\(\frac{1}{16}\)
- \(\frac{1}{8}\)
- \(\frac{1}{4}\)
- \(\frac{1}{2}\)
The correct answer is: B
Explanation
Fraction remaining = (\(\frac{1}{2}\))\(^{\frac{t}{T_{\frac{1}{2}}}}\)
BUT \(\frac{t}{T_{\frac{1}{2}}}\) = \(\frac{80}{20}\) = 4
Fraction remaining = (\(\frac{1}{2})^4\) = \(\frac{1}{16}\)