Iron -59 has a half -life of 45.1 days. How old is an iron nail if the Fe-59 content is 25% that of a new sample of iron? Show all calculations leading to a solution.
First determine the constant, k, for the reaction.
k = 0.693/t1/2
Then ln(No/N) = kt
Let No be 100 (but you may use any number you want as long as N is 25% of No.
Then N = 25
k from above.
Solve for t in days.
Anyone help me through the steps, please?
If you want more help the only thing we can do is to work the problem and present it to you on a platter. And whatever it is that you don't understand gets put off until another day. I have an alternative. Why not explain what you don't understand. The steps are there. There is no chemistry involved. It's all just a little arithmetic. Have you thought of plugging those numbers into the equation and seeing what happens?
Would please work the problem with me? Thanks!
Is Dr. Bob saying
No/N=Kt
After finding the constant (1.38?)=k
100/25=1.38t
4=1.38t
2.89=t ????
That's what I got and it doesn't seem right to me.
To determine the age of the iron nail, we need to use the concept of half-life and the given information. The half-life of Iron-59 (Fe-59) is 45.1 days.
1. Let's start by understanding what half-life means. Half-life is the time it takes for half of a given sample to decay or decrease in quantity. In this case, after one half-life (45.1 days), the Fe-59 will have reduced to 50% of its original quantity.
2. Since we know that the Fe-59 content in the nail is 25% (or 0.25) compared to a new sample of iron, it means that the remaining Fe-59 after one half-life is also 25% (or 0.25).
3. We can use the following formula to calculate the number of half-lives that have occurred:
Remaining fraction = (Initial fraction)^(Number of half-lives)
In this case, the remaining fraction is 0.25, and the initial fraction is also 0.25. Let's denote the number of half-lives as 'n':
0.25 = 0.25^(n)
4. To solve for 'n,' we can take the log of both sides of the equation to remove the exponent:
log(0.25) = log(0.25^n)
Since the left side is a constant, we can simplify the equation:
-0.602 = n * log(0.25)
5. Now, we can solve for 'n' by dividing both sides of the equation by log(0.25):
n = -0.602 / log(0.25)
Using a calculator, we find that n is approximately 2.934.
6. Since 'n' represents the number of half-lives, we can multiply it by the half-life duration (45.1 days) to find the age of the iron nail:
Age = n * half-life
Age ≈ 2.934 * 45.1
Therefore, the age of the iron nail is approximately 132.266 days.