document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=191308>Question 191308</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Radium-226 is a radioactive element with a half-life of 1600 years. <BR>\n" );
document.write( "How much of a 1000 g sample of the element will be present after 6400 years? <BR>\n" );
document.write( "How long will it take for the 1000 gram sample to decay to 1 gram? <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 #143605 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=143605\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Radium-226 is a radioactive element with a half-life of 1600 years. <BR>\n" );
document.write( "How much of a 1000 g sample of the element will be present after 6400 years?<BR>\n" );
document.write( ":<BR>\n" );
document.write( "The half-life decay formula: A = Ao(2^(-t/h))<BR>\n" );
document.write( "where:<BR>\n" );
document.write( "Ao = initial amt<BR>\n" );
document.write( "A = resulting amt after t years<BR>\n" );
document.write( "h = half-life of substance<BR>\n" );
document.write( "t = time in years<BR>\n" );
document.write( ":<BR>\n" );
document.write( "A = 1000(2^(-6400/1600))<BR>\n" );
document.write( "A = 1000(2^-4)<BR>\n" );
document.write( "A = 1000 * .0625<BR>\n" );
document.write( "A = 62.5 grams after 6400 yrs<BR>\n" );
document.write( ":<BR>\n" );
document.write( ":<BR>\n" );
document.write( "How long will it take for the 1000 gram sample to decay to 1 gram?<BR>\n" );
document.write( "1 = 1000(2^(-t/1600))<BR>\n" );
document.write( "Divide both sides by 1000 and we can write this: <BR>\n" );
document.write( "2^(-t/1600) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F1000 ALIGN=MIDDLE  ALT=\"1%2F1000\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "2^(-t/1600) = .001<BR>\n" );
document.write( "log(2^(-t/1600)) = log(.001)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F1600 ALIGN=MIDDLE  ALT=\"-t%2F1600\"   DISABLED_style= \"max-width:90%; height:auto;\" >*log(2) = log(.001)<BR>\n" );
document.write( " <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F1600 ALIGN=MIDDLE  ALT=\"-t%2F1600\"   DISABLED_style= \"max-width:90%; height:auto;\" >*.30103 = -3<BR>\n" );
document.write( "Multiply both sides by 1600, results<BR>\n" );
document.write( "-.30103t = -4800<BR>\n" );
document.write( "t = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-4800%29%2F%28-.30103%29 ALIGN=MIDDLE  ALT=\"%28-4800%29%2F%28-.30103%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "t = 15,945.25 yrs for 1000 g to decay to 1 g<BR>\n" );
document.write( ";<BR>\n" );
document.write( ":<BR>\n" );
document.write( "Check by using this in the following equation with a calc<BR>\n" );
document.write( "A = 1000(2^(-15945.25/1600))<BR>\n" );
document.write( "A = 1.000; confirms our solution   </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );