document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1017011>Question 1017011</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The half-life of plutonium-241 is approximately 13 years.<BR>\n" );
document.write( "a. How much of a sample weighing 3 g will remain after 80 years?<BR>\n" );
document.write( "b. How much time is necessary for a sample weighing 3 g to decay to 0.1g? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #633337 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=633337\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P=pexp(-kt)<BR>\n" );
document.write( "0.5=exp(-kt)<BR>\n" );
document.write( "ln of both sides<BR>\n" );
document.write( "ln2=-13k<BR>\n" );
document.write( "k=1n2/13=0.0533<BR>\n" );
document.write( "Now use that for time=80 years; it is 6 half lives, so I would expect about 1% to be left.<BR>\n" );
document.write( "P=3*exp(-0.0533*80)=3 exp(-4.26558)=3*0.014=0.042 gm, which is about 1%.<BR>\n" );
document.write( "--------------------<BR>\n" );
document.write( "To decay to 0.1 gm<BR>\n" );
document.write( "This should take less time.<BR>\n" );
document.write( "it is 1/30th of the original amount<BR>\n" );
document.write( "ln(1/30)=-3.401<BR>\n" );
document.write( "That equals -0.0533t<BR>\n" );
document.write( "63.8 years.<BR>\n" );
document.write( "Can also do it by <BR>\n" );
document.write( "ln0.1=3exp(-0.0533t)<BR>\n" );
document.write( "and take logs of both sides.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );