A radioactive substance has a half-life of 20 hours. What fraction of the original radioactive nuclide will remain after 80 hours?
1/32
\(\frac{1}{16}\)
\(\frac{1}{8}\)
\(\frac{1}{4}\)
\(\frac{1}{2}\)
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let the original substance = 1. therefore,
1 -. 1/2 - 1/4. - 1/8. -. 1/16
20hrs. 20hrs. 20hrs. 20hrs
Hence, after 80hrs, it remain 1/16 of the original
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let the original substance = 1. therefore,
1 -. 1/2 - 1/4. - 1/8. -. 1/16
20hrs. 20hrs. 20hrs. 20hrs
Hence, after 80hrs, it remain 1/16 of the original
A radioactive substance has a half-life of 20 hours. What traction of the original radioactive nuclide will remain after 80 hours?
A.
1/32
B.
1/16
C.
1/8
D.
¼
E.
1/2
the answer should be 1/8
