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? Please show the calculating so I can learn to do it.
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To find the age of the iron nail, we need to use the concept of radioactive decay and the equation for half-life. The decay of Iron-59 (Fe-59) over time can be described using the following equation:
N(t) = N₀ * (1/2)^(t/t₁/₂)
Where:
- N(t) is the amount of remaining Fe-59 at time 't'
- N₀ is the initial amount of Fe-59
- t is the time that has passed
- t₁/₂ is the half-life of Fe-59
In this case, we know that the content of Fe-59 in the nail is 25% of that in a new sample. This means the remaining amount of Fe-59 (N(t)) in the nail is 0.25 times the initial amount (N₀).
Now, we can substitute the given values into the equation and solve for 't' (time):
N(t) = 0.25 * N₀
(0.25 * N₀) = N₀ * (1/2)^(t/t₁/₂)
To simplify this equation, we can cancel out N₀ on both sides:
0.25 = (1/2)^(t/t₁/₂)
Next, we can take the logarithm (base 2) of both sides of the equation to remove the exponent:
log₂(0.25) = log₂((1/2)^(t/t₁/₂))
Using the property of logarithms, we can bring down the exponent:
log₂(0.25) = (t/t₁/₂) * log₂(1/2)
Since log₂(1/2) equals to -1, we can substitute:
log₂(0.25) = (t/t₁/₂) * (-1)
Now, we can rearrange the equation to solve for 't':
t/t₁/₂ = log₂(0.25)/(-1)
Finally, multiply both sides of the equation by t₁/₂ to isolate 't':
t = (-t₁/₂) * log₂(0.25)
Substituting the given half-life of Iron-59 (t₁/₂ = 45.1 days) into the equation, we can calculate the age of the iron nail.