Question #43a81

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Question #43a81

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Answer 1 · 2015-10-05T20:49:24

$\frac{1}{4096}$ $1/4096$

Explanation:

The idea with nuclear half-life calculations is that the initial mass of the isotope will be halved with every passing of the isotope's half-life.

If you initial amount of fluorine-21 is $A_{0}$ $A_0$ , the you know that you will be left with

$\frac{A_{0}}{2} \to$ $A_0/2 ->$ after one half-life;

What will happen after a second half-life passes?

Well, the amount will be halved again, so you know that you will be left with

$\frac{\frac{A_{0}}{2}}{2} = \frac{A_{0}}{2} \cdot \frac{1}{2} = \frac{A_{0}}{2^{2}} \to$ $(A_0/2)/2 = A_0/2 * 1/2 = A_0/2^2 ->$ after two half-lives

What about after another half-time passes?

$\frac{A_{0}}{4} \cdot \frac{1}{2} = \frac{A_{0}}{2^{3}} \to$ $A_0/4 * 1/2 = A_0/2^3 ->$ after three half-lives

The pattern is very clear at this point - to get the amount of a radioactive isotope that remains after $n$ $color(blue)(n)$ half-lives pass, you need to have

$A_{n} = \frac{A_{0}}{2^{n}}$ $A_n = A_0/2^color(blue)(n)$

In your case, the half-life of fluorine-21 isotope is 5 seconds. How many half-lives must pass for one minute to pass?

$1color(red)(cancel(color(black)("min"))) * (60color(red)(cancel(color(black)("s"))))/(1color(red)(cancel(color(black)("min")))) * "1 half-life"/(5color(red)(cancel(color(black)("s")))) = "12 half-lives"$

This means that after one minute passes you will be left with

$A_"1 minute" = A_0/2^12 = A_0 * 1/4096$

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Question #43a81

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