document.write( "Question 328263: I need on how to solve this. Please show work. Thanks!\r
\n" ); document.write( "\n" ); document.write( "The Radioactive element americium-241 has a half-life of 432 years. Suppose we start with a 45-g mass of americim-241. How much will be left after 329 years? compute the answer to three significant digits. \r
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Algebra.Com's Answer #235178 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Radioactive decay problems are modeled using an exponential function,
\n" ); document.write( "\"N%28t%29=N0%2Ae%5E%28kt%29\"
\n" ); document.write( "where N0 is the initial amount of material, k is a constant for each material.
\n" ); document.write( "Half life means that 1/2 of the initial amount remains.
\n" ); document.write( "We use this to calculate k.
\n" ); document.write( "\"%281%2F2%29N0=N0%2Ae%5E%28k%28432%29%29\"
\n" ); document.write( "\"e%5E%28432k%29=%281%2F2%29\"
\n" ); document.write( "\"432k=ln%281%2F2%29\"
\n" ); document.write( "\"k=ln%281%2F2%29%2F432=-1.605x10%5E%28-3%29\"
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\n" ); document.write( "So now find N(329).
\n" ); document.write( "\"kt=%28329%29%28-1.605x10%5E%28-3%29%29=-0.527883\"
\n" ); document.write( "\"N%28329%29=45%2Ae%5E%28-0.527883%29\"
\n" ); document.write( "\"N%28329%29=45%2A%280.58985%29\"
\n" ); document.write( "\"N%28329%29=26.543\"grams
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