Question

Suppose the isotope your skittles® represented was uranium-238 and the trials represent the number of half-lives. how old was the sample at the end of your tests? use table 1 in the introduction to help you answer this question. include your calculations. 3. if 1/8 of a radioactive element remains after 600 years, what is that element's half-life? 4. identify and describe similarities and differences between this experiment and radioactive decay in nature.

Answer 1

The age of a sample after a certain number of half-lives can be calculated, and the half-life of a radioactive element can be determined if the fraction remaining after a given time is known. The predictable rate of decay is a similarity between an experiment and natural radioactive decay; the presence of decay products in nature is a difference.

Step-by-step explanation:

The age of a sample can be determined using the concept of half-lives. For example, if 1/8 of a radioactive element remains after 600 years, it means that three half-lives have passed (as 1/2 to the power of 3 equals 1/8). You can calculate the half-life by dividing the total time elapsed by the number of half-lives that have passed. Therefore, the half-life would be 600 years / 3, which equals 200 years.

Similarities between this experiment and radioactive decay in nature include the predictable rate at which radioactive isotopes decay and the fact that this process is not influenced by external conditions, like temperature or pressure. A difference is that in nature, decay products can also be radioactive and continue the decay chain, which isn't typically part of a controlled classroom experiment.

Answer 2

I can't answer the first and the last question because no given data is provided. However, I can answer the second question.

if 1/8 of a radioactive element remains after 600 years, what is that element's half-life?

The formula for half life, t1/2, is:
t1/2 = ln2/K

So we must determine K first. The formula for that is:
A = A₀e^(-kt)
where
A = 1/8*A₀
t = 600 years
Thus,
1/8*A₀ = A₀e^(-k*600)
1/8 = e^(-600k)
Solving for k,
K = 3.466×10⁻³

Therefore,
t1/2 = ln2/3.466×10⁻³ = 200 years