document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=635350>Question 635350</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Ten grams of Uranium will decay to 2.5 g in 496,000 years. What is its half-life?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #400265 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=400265\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Ten grams of Uranium will decay to 2.5 g in 496,000 years. What is its half-life?<BR>\n" );
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document.write( "Formula for radioactive decay: A/Ao=2^-t/h, A=radioactive material present at time t, Ao=amt present initially, h=material half life.<BR>\n" );
document.write( "2.5/10=2^-4.96*10^5/h<BR>\n" );
document.write( ".25=2^-4.96*10^5/h<BR>\n" );
document.write( "take log of both sides<BR>\n" );
document.write( "log.25=-4.96*10^5/h)log2<BR>\n" );
document.write( "h(log.25)=-(4.96*10^5)log2<BR>\n" );
document.write( "h=-(4.96*10^5)log2/log.25=-2.48*10^5<BR>\n" );
document.write( "..<BR>\n" );
document.write( "half-life of material=2.48*10^5 yrs </SPAN>\n" );
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