document.write( "Question 378561: The radioactive element americium-241 has half the life of 432 years. Suppose we start with a 45-g mass of americium-241, how much would be left after 343 years? \n" ); document.write( "
Algebra.Com's Answer #268892 by richard1234(7193)\"\" \"About 
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The formula for determining the amount of substance left is \"A+=+Ce%5E%28kt%29\", where C is the initial amount, k is the decay constant, and t is time. We must solve for k:\r
\n" ); document.write( "\n" ); document.write( "Given the half-life of 432 years, we may plug in t:\r
\n" ); document.write( "\n" ); document.write( "\"C%2F2+=+Ce%5E%28432k%29\"\r
\n" ); document.write( "\n" ); document.write( "\"1%2F2+=+e%5E%28432k%29\"\r
\n" ); document.write( "\n" ); document.write( "Taking ln of both sides,\r
\n" ); document.write( "\n" ); document.write( "\"ln+%281%2F2%29+=+432k\"\r
\n" ); document.write( "\n" ); document.write( "\"k+=+%28ln+%281%2F2%29%29%2F432\" = -0.00160...\r
\n" ); document.write( "\n" ); document.write( "Now, we substitute t = 343 years to our equation to obtain\r
\n" ); document.write( "\n" ); document.write( "\"A+=+45%2Ae%5E%28343%2A-.00160%29\" = 25.95376 (grams)
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