A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?
\( \frac{1}{32} \)
\( \frac{1}{16} \)
\( \frac{1}{8} \)
\( \frac{1}{4} \)
Correct answer is B
80 days = 4 half lives (i.e 4 x 20dqs = 80dq)
But the fraction always left by any radioactive material after 4 half lives is
\( \left ( \frac{1}{2} \right)^4 = \frac{1}{16} \)