Question

The radioactive substance uranium-240 has a half-life of 14 hours. The amount At of a sample of uranium-240 remaining (in grams) after t hours is given by the following exponential function.

Answer 1

Step 1

Given;

$A(t)=5600(\frac{1}{2})^{\frac{t}{14}}$

Required; To find the amount A(t) of a sample of uranium-240 remaining (in grams) after 13 hours and 60 hours

Step 2

Find the initial amount. To do this we set t=0

$\begin{gathered} A(0)=5600(\frac{1}{2}_)^{\frac{0}{14}} \\ A(0)=5600grams \end{gathered}$

Hence the equation remains valid

Step 3

Find the amount A(t) left after 13 hours

$A(13)=5600(\frac{1}{2})^{\frac{13}{14}}$$\begin{gathered} =5600\cdot \frac{1^{\frac{13}{14}}}{2^{\frac{13}{14}}} \\ =\frac{5600}{1}\cdot \frac{1}{2^{\frac{13}{14}}} \\ =\frac{5600}{2^{\frac{13}{14}}} \\ =\frac{2^5\cdot \:175}{2^{\frac{13}{14}}} \\ =2^{\frac{57}{14}}\cdot\:175=2942.11858 \\ =2942grams \end{gathered}$

Step 4

Find the amount A(t) left after 60 hours

$\begin{gathered} A(60)=5600(\frac{1}{2})^^{\frac{60}{14}} \\ =5600\cdot \frac{1^{\frac{60}{14}}}{2^{\frac{60}{14}}} \\ =5600\cdot \frac{1}{2^{\frac{60}{14}}} \\ =\frac{5600}{1}\cdot \frac{1}{16\cdot \:2^{\frac{2}{7}}} \\ =\frac{5600}{16\cdot \:2^{\frac{2}{7}}} \\ =\frac{16\cdot \:350}{16\cdot \:2^{\frac{2}{7}}} \\ =\frac{350}{2^{\frac{2}{7}}} \\ =2^{\frac{5}{7}}\cdot \:175 \\ A(60)=287.11737 \\ A(60)\approx287grams \end{gathered}$

Answers;

$\begin{gathered} A(13)=2942grams\text{ to the nearest gram} \\ A(60)=287grams\text{ to the nearest gram} \end{gathered}$