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

Guest Feb 4, 2015

#2**+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**

Melody
Feb 4, 2015

#2**+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