NEED SOME SERIOUS CHEMISTRY HELP PEOPLE. 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 this answer.
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A is the amount after decay.
Ao is the initial amount.
e is a number
k is a constant based on the radioactive material
t is time
Does that make sense so far?
Basically you don't need too much calculations. By definition in one half life you will have 50% of the original amount of sample and in two half life you will have the 50% of the 50% of the original sample. In other words you will have the 25% of the original sample. Then after two half life you will have the 25% of the original sample, and if the half life of this isotope Fe-59 is 45.1 days, then after (45.1days x 2) you will have the 25% of your sample. That is how old the iron nail is, not too old just a few months.
ahhhhhh Thank you!
Please can you do all the calculation for this question please.
Use The formula From @JoannaBlackweler posted above an follow the procedures described in this old Yahoo! Post
chemistry - bobpursley, Monday, May 11, 2015 at 2:08pm
25=100 e^(-.693 t/45.1)
take the ln of each side...
Ln(25)=ln(100) -.693 t/45.1
t= - 45.1/.693 ( ln .25)
putting that into the goggle calculator..
Now the easy way. to reduct to 1/4, it takes two half lives, or you do it.
I did nit understand this please can some one help me to understand? Where come -.693???
that number comes from the linearization of the first order reaction and it is equal to ln2 (natural logarithm of 2) because when you introduce in the formula the half life has 1/2 and then apply logarithm you get in the formula ln 2.
Also you can think about it like the number Pi in the formula to calculate the length of the circumference. \[2\pi r\]