document.write( "Question 122244: A radioactive isotope has a half-life of 3000 years. If a sample of this isotope origanally has a mass 30g, what equation would model the mass of this sample over time? What would its mass be after 5 hours? \n" ); document.write( "
Algebra.Com's Answer #89739 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
A radioactive isotope has a half-life of 3000 years. If a sample of this isotope origanally has a mass 30g, what equation would model the mass of this sample over time? What would its mass be after 5 hours?
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document.write( "Let A(t) represent the amount of \r\n" );
document.write( "isotope at time t. The formula is:\r\n" );
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document.write( "A(t) = Pert\r\n" );
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document.write( "At the start, when time = t = 0, \r\n" );
document.write( "then A(0) = 30\r\n" );
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document.write( "So we substitute 0 for t:\r\n" );
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document.write( "A(0) = Per*0\r\n" );
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document.write( "and we substitute 30 for A(0)\r\n" );
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document.write( "  30 = Pe0\r\n" );
document.write( "  30 = P(1)\r\n" );
document.write( "  30 = P\r\n" );
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document.write( "So now the formula\r\n" );
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document.write( "A(t) = Pert\r\n" );
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document.write( "becomes\r\n" );
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document.write( "A(t) = 30ert\r\n" );
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document.write( "Nw we read >>...half-life of 3000 years...<<\r\n" );
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document.write( "which means that in 3000 years the isotope will\r\n" );
document.write( "have reduced to half of its original mass of \r\n" );
document.write( "30 grams.  That means that A(3000) = 15 grams.\r\n" );
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document.write( "So we substitute 3000 for t\r\n" );
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document.write( "A(t) = 30ert\r\n" );
document.write( "A(3000) = 30er(3000)\r\n" );
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document.write( "and then substitute 15 for A(3000)\r\n" );
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document.write( "A(3000) = 30er(3000)\r\n" );
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document.write( "15 = 30er(3000)\r\n" );
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document.write( "Divide both sides by 30\r\n" );
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document.write( "\"15%2F30\" = e3000r\r\n" );
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document.write( "0.5 = e3000r\r\n" );
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document.write( "Taking the natural log of both sides,\r\n" );
document.write( "we have \r\n" );
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document.write( "ln(0.5) = 3000r\r\n" );
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document.write( "Use a calculato to get the left side:\r\n" );
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document.write( "-.6931471806 = 3000r\r\n" );
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document.write( "Divide both sides by 3000\r\n" );
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document.write( "-.00023104906 = r\r\n" );
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document.write( "So the formula\r\n" );
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document.write( "A(t) = 30ert\r\n" );
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document.write( "becomes\r\n" );
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document.write( "A(t) = 30e-0.00023104906t\r\n" );
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document.write( "That is the equation that models the mass,\r\n" );
document.write( "which was the first part of your problem. \r\n" );
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document.write( "Now for the second part, we only need to plug \r\n" );
document.write( "in 5 hours.  But we must change that to\r\n" );
document.write( "years.  So\r\n" );
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document.write( "Change 5 hours to days by dividing by 24, getting\r\n" );
document.write( "5 hours = 0.2083333333 days,\r\n" );
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document.write( "Now divide that by 365.25 to change it to years:\r\n" );
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document.write( "5 hours = 0.0005703855807 years\r\n" );
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document.write( "So plug that into\r\n" );
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document.write( "A(t) = 30e-.00023104906t\r\n" );
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document.write( "A(0.0005703855807) = 30e-.00023104906(0.0005703855807)\r\n" );
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document.write( "A(0.0005703855807) = 29.99999605 grams.\r\n" );
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document.write( "As we might guess, if it takes 3000 years for\r\n" );
document.write( "30 grams to decrease to 15 grams, we wouldn't\r\n" );
document.write( "expect it to have noticeably decreased from the\r\n" );
document.write( "original 30 grams after only 5 hours!\r\n" );
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document.write( "Edwin

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