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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?

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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

#2
+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
0

1.25 grams i think

Feb 4, 2015
#2
+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

Melody Feb 4, 2015