document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=673868>Question 673868</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The radioactive element americium-241 has a half-life of 432 years. Suppose we start with a 20-g mass of americium-241.<BR>\n" );
document.write( "How much will be left after 367 years? Compute the answer to three significant digits.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #419082 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=419082\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  The radioactive element americium-241 has a half-life of 432 years.<BR>\n" );
document.write( " Suppose we start with a 20-g mass of americium-241.<BR>\n" );
document.write( "How much will be left after 367 years? <BR>\n" );
document.write( ":<BR>\n" );
document.write( "the radioactive decay formula: A = Ao*2^(-t/h), where:<BR>\n" );
document.write( "A = resulting amt after t time<BR>\n" );
document.write( "Ao = initial amt (t=0)<BR>\n" );
document.write( "t = time of decay<BR>\n" );
document.write( "h = half-life of substance<BR>\n" );
document.write( ":<BR>\n" );
document.write( "A = 20*2^(-367/432)<BR>\n" );
document.write( "Use a calc<BR>\n" );
document.write( "A = 20 * .5549628<BR>\n" );
document.write( "A = 11.099 grams after 367 yrs </SPAN>\n" );
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