document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1196893>Question 1196893</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The radioactive isotope carbon 14 used to date fossils decays with an annual rate of about 0.000124. If a fossil is found which originally has 2 mg of carbon 14, and it now has 0.18 mg, how old is it? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #829934 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=829934\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Some comments about the problem and the other responses you have received....<br><BR>\n" );
document.write( "(1) Your decay rate of 0.000124 corresponds to a carbon 14 half life of 5590 years.  Internet sources place the half life at 5730 plus or minus 30 years.<br><BR>\n" );
document.write( "(2) Radioactive decay is a statistical process; the amount remaining does not follow a smooth decaying exponential curve.  Any age obtained using carbon 14 dating can only be treated as approximate.  Any answer that gives a carbon 14 age should show no more than about 3 significant figures.<br><BR>\n" );
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