document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=379009>Question 379009</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Plutonium 239 is used as fuel for some nuclear reactors, and also as the fissionable material in atomic bombs. it has a half life of 24400 years. how long would it take 10 grams of plutonium 239 to decay to 1 gram?(round your answer to three significant digits) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #269264 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=269264\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Plutonium 239 is used as fuel for some nuclear reactors, and also as the<BR>\n" );
document.write( " fissionable material in atomic bombs.<BR>\n" );
document.write( " it has a half life of 24400 years. how long would it take 10 grams of<BR>\n" );
document.write( " plutonium 239 to decay to 1 gram?(round your answer to three significant digits)<BR>\n" );
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document.write( "The half life formula: Ao*2^(-t/h) = A<BR>\n" );
document.write( "where<BR>\n" );
document.write( "A = resulting amt after t yrs<BR>\n" );
document.write( "Ao = initial amt<BR>\n" );
document.write( "t = time in yrs<BR>\n" );
document.write( "h = half-life of substance<BR>\n" );
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document.write( "10*2(-t/24400) = 1<BR>\n" );
document.write( "Divide both sides by 10<BR>\n" );
document.write( "2(-t/24400) = .1<BR>\n" );
document.write( "Using nat logs<BR>\n" );
document.write( "ln(2(-t/24400)) = ln(.1)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F24400 ALIGN=MIDDLE  ALT=\"-t%2F24400\"   DISABLED_style= \"max-width:90%; height:auto;\" >ln(2) = ln(.1)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F24400 ALIGN=MIDDLE  ALT=\"-t%2F24400\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%28.1%29%2Fln%282%29 ALIGN=MIDDLE  ALT=\"ln%28.1%29%2Fln%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Using a calc<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-t%2F24400 ALIGN=MIDDLE  ALT=\"-t%2F24400\"   DISABLED_style= \"max-width:90%; height:auto;\" > = -3.32193<BR>\n" );
document.write( "Multiply both sides by -24400<BR>\n" );
document.write( "t = 81,055.0456 years<BR>\n" );
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document.write( "Check solution on a calc; enter 10*2^(-81055/24400) results: 1.000<BR>\n" );
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