There are 10 grams of Curium-245 which has a half-life of 9300 years. How many grams will remain after 37200 years?
+118608 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
+118608 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
