document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=428542>Question 428542</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose you begin with 2.8 mg of a pure radioactive substance and 3.6 h later determine that 0.35 mg remain. What is the half-life of the substance? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #297840 by </B><A HREF=/tutors/aboutme.mpl?userid=Gogonati><B>Gogonati(855)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Gogonati><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=Gogonati><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=297840\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Solution:As we know the radioactive decay is given by: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=M%28t%29=Mo%2Ae%5E%28-k%2At%29 ALIGN=MIDDLE  ALT=\"M%28t%29=Mo%2Ae%5E%28-k%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( " where M is th actual mass of the substance, Mo is the mass after t= 3.6  h, <BR>\n" );
document.write( "   and k is decay coefficient. Substitute our values:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0.35=2.8%2Ae%5E-3.6k ALIGN=MIDDLE  ALT=\"0.35=2.8%2Ae%5E-3.6k\"   DISABLED_style= \"max-width:90%; height:auto;\" >, Take ln for both sides.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%280.35%29=-3.6k%2Aln%28e%29 ALIGN=MIDDLE  ALT=\"ln%280.35%29=-3.6k%2Aln%28e%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "k=-0.212 Now that, we know the coefficient of half-life, can compute the half-life.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0.35=2.8%2Ae%5E%28-0.212%2At%29 ALIGN=MIDDLE  ALT=\"0.35=2.8%2Ae%5E%28-0.212%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, Take ln for both sides.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ln%280.35%29=-0.212t%2Alne ALIGN=MIDDLE  ALT=\"ln%280.35%29=-0.212t%2Alne\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=t=4.8 ALIGN=MIDDLE  ALT=\"t=4.8\"   DISABLED_style= \"max-width:90%; height:auto;\" > h.  </SPAN>\n" );
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