document.write( "Question 252184: Beryllium-11 decomposes into boron-11 with a half-life of 13.8 secs. How long will it take 240g of beryllium-11 to decompose into 7.5g of beryllium-11? \n" ); document.write( "

Algebra.Com's Answer #183954 by nerdybill(7384) $\"\"$ $\"About$

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Beryllium-11 decomposes into boron-11 with a half-life of 13.8 secs. How long will it take 240g of beryllium-11 to decompose into 7.5g of beryllium-11?
\n" ); document.write( ".
\n" ); document.write( "Apply the exponential growth formula:
\n" ); document.write( "x(t) = xo*e^(kt)
\n" ); document.write( "where
\n" ); document.write( "x(t) is amount at time t
\n" ); document.write( "xo is the initial amount
\n" ); document.write( "k is the growth/decay rate
\n" ); document.write( "t is the time
\n" ); document.write( ".
\n" ); document.write( "We need to find k, do this with the info provided in the first sentence:
\n" ); document.write( "x(t) = xo*e^(kt)
\n" ); document.write( "x/2 = x*e^(k*13.8)
\n" ); document.write( "1/2 = e^(k*13.8)
\n" ); document.write( ".5 = e^(k*13.8)
\n" ); document.write( "ln(.5) = k*13.8
\n" ); document.write( "ln(.5)/13.8 = k
\n" ); document.write( ".
\n" ); document.write( "Use the above to answer the question:
\n" ); document.write( "x(t) = xo*e^(kt)
\n" ); document.write( "7.5 = 240*e^(t*ln(.5)/13.8)
\n" ); document.write( "7.5/240 = e^(t*ln(.5)/13.8)
\n" ); document.write( "ln(7.5/240) = t*ln(.5)/13.8
\n" ); document.write( "[ln(7.5/240)]/[ln(.5)/13.8] = t
\n" ); document.write( "114.84 secs = t
\n" ); document.write( " \n" ); document.write( "

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