document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1023633>Question 1023633</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Using A=Pert formula, On March 11th 2011 a Fukushima reactor released 4.2 times the amount of cesium-137 as was leaked during 1986 Chernobyl disaster.<BR>\n" );
document.write( "*half-life of cesium-137 is 30.2 years.<BR>\n" );
document.write( "a.In what year will the cesium-137 level be half as much as immediately after the 2011 Fukushima disaster.<BR>\n" );
document.write( "b.In what year will the cesium-137 level be a quarter as much immediately after 2011 Fukushima disaster?<BR>\n" );
document.write( "c. How long after the disaster until the cesium-137 level is the same as immediately after the 1986 Chernobyl disaster.<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 #692336 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel1380><B>jorel1380(3719)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel1380><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=jorel1380><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=692336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a.1/2=e^-30.2r<BR>\n" );
document.write( "-ln 2=ln e^-30.2r<BR>\n" );
document.write( "-0.69314718055994530941723212145818=ln e^-30.2r=-30.2r<BR>\n" );
document.write( "r=0.02295189339602467911977589806153<BR>\n" );
document.write( "The rate of decay constant is 0.022951893. 2011+30.2=2041 as the year in which half of the cesium is gone.\r<BR>\n" );
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document.write( "b.The amount of cesium left after t is 1/4. So:<BR>\n" );
document.write( "ln 1/4=-0.022951893t<BR>\n" );
document.write( "ln 1/4=-ln 4=-1.3862943611198906188344642429164<BR>\n" );
document.write( "t=-1.3862943611198906188344642429164/-0.022951893=60.40000 years.<BR>\n" );
document.write( "2011+60.4=2071 as the year cesium levels are 1/4 of what they were in 2011.\r<BR>\n" );
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document.write( "c.Since the reactor released 4.2 times the amount released in Chernobyl, we have:<BR>\n" );
document.write( "ln 1/4.2=-0.022951893t<BR>\n" );
document.write( "-ln 4.2=-0.022951893t<BR>\n" );
document.write( "t=62.525758781 years after 2011, or 2073.52 when Fukushima levels equal Chernobyl levels.  </SPAN>\n" );
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