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There are 10 grams of Curium-245 which has a half-life of 9300 years. How many grams will remain after 37200 years?

 Feb 4, 2015

Best Answer 

 #2
avatar+94976 
+5

There are 10 grams of Curium-245 which has a half-life of 9300 years. How many grams will remain after 37200 years?

 

$$\begin{array}{rll}
A&=&A_0e^{kt}\\\\
M&=&2Me^{k*9300}\\\\
0.5&=&e^{k*9300}\\\\
ln(0.5)&=&ln(e^{k*9300})\\\\
ln(0.5)&=&(k*9300)ln(e)\\\\
ln(0.5)&=&k*9300\\\\
k&=&\frac{ln0.5}{9300}\\\\
So&&\\\\
A&=&A_0e^{\frac{ln0.5}{9300}*t}\\\\
&&$Start with 10g. How many grams will remain after 37200 years?$\\\\
A&=&10e^{\frac{ln0.5}{9300}*37200}\\\\
\end{array}$$

 

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\frac{{{log}}_{{\mathtt{e}}}{\left({\mathtt{0.5}}\right)}}{{\mathtt{9\,300}}}}{\mathtt{\,\times\,}}{\mathtt{37\,200}}\right)} = {\frac{{\mathtt{5}}}{{\mathtt{8}}}} = {\mathtt{0.625\: \!000\: \!000\: \!000\: \!000\: \!1}}$$

 

so        625mg will remain

 Feb 4, 2015
 #1
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0

1.25 grams i think

 Feb 4, 2015
 #2
avatar+94976 
+5
Best Answer

There are 10 grams of Curium-245 which has a half-life of 9300 years. How many grams will remain after 37200 years?

 

$$\begin{array}{rll}
A&=&A_0e^{kt}\\\\
M&=&2Me^{k*9300}\\\\
0.5&=&e^{k*9300}\\\\
ln(0.5)&=&ln(e^{k*9300})\\\\
ln(0.5)&=&(k*9300)ln(e)\\\\
ln(0.5)&=&k*9300\\\\
k&=&\frac{ln0.5}{9300}\\\\
So&&\\\\
A&=&A_0e^{\frac{ln0.5}{9300}*t}\\\\
&&$Start with 10g. How many grams will remain after 37200 years?$\\\\
A&=&10e^{\frac{ln0.5}{9300}*37200}\\\\
\end{array}$$

 

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\frac{{{log}}_{{\mathtt{e}}}{\left({\mathtt{0.5}}\right)}}{{\mathtt{9\,300}}}}{\mathtt{\,\times\,}}{\mathtt{37\,200}}\right)} = {\frac{{\mathtt{5}}}{{\mathtt{8}}}} = {\mathtt{0.625\: \!000\: \!000\: \!000\: \!000\: \!1}}$$

 

so        625mg will remain

Melody Feb 4, 2015

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