Question

120 atoms: 40 have decayed, what is the half life? What is it in years?

Answer 1

Answer on Question 57424, Physics, Other

Question:

120 atoms: 40 have decayed, what is the half-life? What is it in years?

Solution:

Given: beginning amount of atoms – 120 atoms; 40 have decayed means the ending amount of atoms is $120 - 40 = 80$ atoms.

Let’s first calculate number of half-lives, $n$ from the formula:

$$Ending\ Amount = \frac{Beginning\ Amount}{2^n},$$ $$\left(\frac{1}{2}\right)^n = \frac{Ending\ Amount}{Beginning\ Amount} = \frac{80}{120}.$$

Let’s take the $log$ of both sides of equation:

$$\log \left(\frac{1}{2}\right)^n = \log \left(\frac{80}{120}\right),$$ $$n \cdot \log(0.5) = \log \left(\frac{80}{120}\right),$$ $$n = \log \left(\frac{80}{120}\right) / \log(0.5) = 0.585.$$

In order to find the half-life we must know the elapsed time. We can find it from the formula:

$$Beginning\ Amount \cdot \left(\frac{1}{2}\right)^{\left(\frac{Elaps.time}{n}\right)} = Ending\ Amount,$$ $$120 \cdot \left(\frac{1}{2}\right)^{\left(\frac{Elaps.time}{0.585}\right)} = 80,$$ $$\left(\frac{1}{2}\right)^{\left(\frac{Elaps.time}{0.585}\right)} = \frac{80}{120}.$$

Again take the $log$ of both sides of equation:

$$\log \left(\frac {1}{2}\right) ^ {\left(\frac {\text {Elaps.time}}{0 . 5 8 5}\right)} = \log \left(\frac {8 0}{1 2 0}\right),$$ $$\log (0. 5) \cdot \left(\frac {\text {Elaps. time}}{0 . 5 8 5}\right) = \log \left(\frac {8 0}{1 2 0}\right),$$ $$E l a p s. t i m e = 0. 5 8 5 \cdot \frac {\log \left(\frac {8 0}{1 2 0}\right)}{\log (0 . 5)} = 0. 3 4 2 y e a r.$$

Then we can find the half-life from the formula:

$$T _ {1 / 2} = \frac {\text {Elaps. time} \cdot \log 2}{\log \left(\frac {\text {Beginning Amount}}{\text {Ending Amount}}\right)} = \frac {0 . 3 4 2 y e a r \cdot \log 2}{\log \left(\frac {8 0}{1 2 0}\right)} = 0. 5 8 4 y e a r.$$

Answer:

$$T _ {1 / 2} = 0. 5 8 4 y e a r.$$

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