SOLUTION: Gallium-67 is used in nuclear medicine to help doctors locate inflammation and chronic infections. The patient is injected with a tracer (trace amount) that includes gallium-67, wh
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Question 1194724: Gallium-67 is used in nuclear medicine to help doctors locate inflammation and chronic infections. The patient is injected with a tracer (trace amount) that includes gallium-67, which collects in areas of inflammation and infection. The gallium-67 emits radiation that a special camera can detect. Gallium-67 has a half-life of 3.26 days.
- Give an exponential equation to represent the percentage of the original gallium-67 after t days.
- Determine the amount of gallium-67 left after 4 days.
-Solve your equation to determine the time it will take for there to be 1% of the original gallium-67.
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Half-life of Gallium-67 is 3.26 days.
Revised decay equation:
time t in days
p original amount
A amount after time t
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
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.
As a mathematician, I find it far easier to use the definition of half life directly.
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
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
ANSWER 1: (converted to a percent)
Plug in t=4 days to get
ANSWER 2: = 42.74 percent
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%.
You can find the answer to the last question by solving the equation
using logarithms.
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 and .
ANSWER 3: 21.66 days
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.
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