Question 1194724
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In my experience, scientists like to use an exponential equation using the natural base e to solve problems involving the half life of radioactive substances -- as shown by the other tutor.<br>
As a mathematician, I find it far easier to use the definition of half life directly.<br>
The half life of Gallium-67 is 3.26 days.  When time is measured in half lives, t days is t/3.26 half lives.  Since 1/2 of the original material is left after one half life, the fraction of the original amount remaining after n half lives is {{{(1/2)^n}}}<br>
In this example, with the tracer element having a half life of 3.26 days, an equation for the fraction remaining after t days is<br>
ANSWER 1: {{{A=A(0)(1/2)^(t/3.26)}}} (converted to a percent)<br>
Plug in t=4 days to get<br>
ANSWER 2: {{{(1/2)^(4/3.26)=0.4272}}} = 42.74 percent<br>
Note that result makes sense; 4 days is a bit more than one half life, so the amount remaining should be a bit less than 50%.<br>
You can find the answer to the last question by solving the equation<br>
{{{(1/2)^(t/3.26)=.01}}}<br>
using logarithms.<br>
But finding an accurate numerical answer will require a calculator; so you might as well just use a graphing calculator to find the intersection of the graphs of {{{y=(1/2)^(t/3.26)}}} and {{{y=0.01}}}.<br>
ANSWER 3: 21.66 days<br>
Note again this result make sense also.  The amount remaining after 6 half lives would be 1/2^6=1/64; the amount remaining after 7 half lives would be 1/2^7=1/128.  1% is 1/100, so the answer should be between 6 and 7 half lives, which is roughly between 19 and 23 days.<br>