Select the correct answer.
The isotope calcium-41 decays into potassium-41, with a half-life of [tex]$1.03 \times 10^5$[/tex] years. There is a sample of calcium-41 containing [tex]$5 \times 10^9$[/tex] atoms. How many atoms of calcium-41 and potassium-41 will there be after [tex]$4.12 \times 10^5$[/tex] years?
A. [tex]$3.125 \times 10^8$[/tex] atoms of calcium-41 and [tex]$4.375 \times 10^9$[/tex] atoms of potassium-41
B. [tex]$6.25 \times 10^6$[/tex] atoms of calcium-41 and [tex]$4.6875 \times 10^9$[/tex] atoms of potassium-41
C. [tex]$6.25 \times 10^8$[/tex] atoms of calcium-41 and [tex]$4.375 \times 10^9$[/tex] atoms of potassium-41
D. [tex]$3.125 \times 10^6$[/tex] atoms of calcium-41 and [tex]$4.6875 \times 10^9$[/tex] atoms of potassium-41
Answer (2)
Calculate the number of half-lives: n = 1.03 × 1 0 5 4.12 × 1 0 5 = 4 .
Calculate the remaining Calcium-41 atoms: N Ca-41 = 5 × 1 0 9 × ( 2 1 ) 4 = 3.125 × 1 0 8 .
Calculate the formed Potassium-41 atoms: N K-41 = 5 × 1 0 9 − 3.125 × 1 0 8 = 4.6875 × 1 0 9 .
The final answer is: 3.125 × 1 0 8 atoms of Calcium-41 and 4.6875 × 1 0 9 atoms of Potassium-41 .
Explanation
Problem Analysis We are given that Calcium-41 decays into Potassium-41 with a half-life of 1.03 × 1 0 5 years. We start with 5 × 1 0 9 atoms of Calcium-41 and want to find out how many atoms of Calcium-41 and Potassium-41 are present after 4.12 × 1 0 5 years.
Calculating Number of Half-Lives First, we need to determine how many half-lives have passed during the given time period. We can calculate this by dividing the total time elapsed by the half-life: n = half-life time elapsed = 1.03 × 1 0 5 4.12 × 1 0 5 = 4
So, 4 half-lives have passed.
Calculating Remaining Calcium-41 Atoms Next, we calculate the number of Calcium-41 atoms remaining after 4 half-lives. The formula for radioactive decay is: N ( t ) = N 0 × ( 2 1 ) n
where N ( t ) is the number of atoms remaining after time t , N 0 is the initial number of atoms, and n is the number of half-lives. In our case, N 0 = 5 × 1 0 9 and n = 4 . Therefore, N Ca-41 = 5 × 1 0 9 × ( 2 1 ) 4 = 5 × 1 0 9 × 16 1 = 0.3125 × 1 0 9 = 3.125 × 1 0 8 So, after 4 half-lives, there are 3.125 × 1 0 8 atoms of Calcium-41 remaining.
Calculating Potassium-41 Atoms Now, we calculate the number of Potassium-41 atoms formed. Since the Calcium-41 atoms decay into Potassium-41, the number of Potassium-41 atoms is the difference between the initial number of Calcium-41 atoms and the remaining number of Calcium-41 atoms: N K-41 = N 0 − N Ca-41 = 5 × 1 0 9 − 3.125 × 1 0 8 = 5 × 1 0 9 − 0.3125 × 1 0 9 = 4.6875 × 1 0 9 So, there are 4.6875 × 1 0 9 atoms of Potassium-41 formed.
Final Answer Therefore, after 4.12 × 1 0 5 years, there will be 3.125 × 1 0 8 atoms of Calcium-41 and 4.6875 × 1 0 9 atoms of Potassium-41. Comparing this to the answer choices, we see that option A matches our result.
Examples
Radioactive decay is used in carbon dating to determine the age of ancient artifacts. By measuring the amount of carbon-14 remaining in an artifact, scientists can estimate how long ago the organism died. This technique relies on the predictable decay rate of radioactive isotopes, similar to the calcium-41 decay in this problem. Understanding radioactive decay helps archaeologists and paleontologists uncover the history of life on Earth.
After 4.12 x 10^5 years, there will be 3.125 x 10^8 atoms of calcium-41 and 4.6875 x 10^9 atoms of potassium-41. The correct answer is Option A.
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