How long will it take for 25% of the C-14 atoms in a sample of C-14 to decay? The half-life for the radioactive decay of C-14 is 5730 years.
If a sample of C-14 initially contains 1.1 mmol of C-14 , how many millimoles will be left after 2275 years?
the answer is 2400 years, based on the equation ln(100/75)/(1.2xE^-4). And the 1.2xE^-4 is the k
Both problems are similar. Use this first equation to evaluate k.
k = 0.693/t1/2
Then substitute k into this equation.
ln(No/N) = kt
To make thing easy I would assume we start with 100 atoms so No = 100. If 25% of the atoms decay, that will leave 75 atoms remaining; therefore, N = 75. You know k from the first equation, you can solve for time.
Second problem is done the same way but you solve for N.
To determine how long it will take for 25% of the C-14 atoms in a sample to decay, we need to use the concept of half-life. The half-life of C-14 is given as 5730 years, which means that after 5730 years, only 50% of the original C-14 atoms will remain.
To find the time it takes for 25% of the C-14 atoms to decay, we can calculate the number of half-lives needed. Since 50% remaining is considered one half-life, 25% remaining would be half of that or 12.5%.
To calculate the number of half-lives needed, we can use the formula:
Number of half-lives = log(remaining percentage) / log(0.5)
Number of half-lives = log(0.125) / log(0.5) ≈ 3.585
Therefore, it will take approximately 3.585 half-lives for 25% of the C-14 atoms to decay.
Now let's calculate the number of millimoles of C-14 that will be left after 2275 years. Here's how you do it:
1. Determine the number of half-lives that have elapsed:
Number of half-lives = Time elapsed / C-14 half-life
Number of half-lives = 2275 years / 5730 years ≈ 0.397 half-lives
2. Calculate the remaining percentage using the formula:
Remaining percentage = (1/2)^(number of half-lives)
Remaining percentage = (1/2)^(0.397) ≈ 0.7986 ≈ 79.86%
3. Calculate the number of millimoles remaining:
Millimoles remaining = Initial millimoles * Remaining percentage
Millimoles remaining = 1.1 mmol * 0.7986 ≈ 0.879 mmol
Therefore, after 2275 years, approximately 0.879 millimoles of C-14 will be left in the sample.