If a radioactive isotope has a half life of 1000 years how long does it take for 3/4 of the original sample to decompose?

1 Answer

If you begin with a 1 gram sample of the isotope, at the end of 1000 years you would have

1 gram x #(1/2)# = #1/2# gram would remain

At the end of the 2nd 1000 years you would

#(1/2) gram # x #(1/2)# = #1/4# gram would remain

Since you began with 1 gram and after 2000 year you would have

#1/4# of a gram remaining. This means #3/4# of the original amount would have decayed.

1 gram - #1/4# gram = #3/4# gram

The answer therefore is 2000 years.

I hope this was helpful.
SMARTERTEACHER

Also, if you are more of a mathematical person, you can use the equation: m=ca^(t/h) where;

"m" is the final mass of the sample
"c" is the starting mass of the sample
"a" is how the substance is decaying (half life = multiplying by 1/2)
"t" is the time it is alive
"h" is the half life time.

For example, if I said: "If you begin with a radioactive isotope weighing 100g and it has a half life of 1000 years, how long does it take to decay to 3/4 (25g) of it's original mass?"

In this case, you know that the final mass, "m", is 25g, you know that the starting mass, "c" is 100g, "a" is 1/2 (it is halving), "t" is the time it is alive (you don't know this), and you know the half life time, "h", is 1000 years. Knowing these, you sub them into the equation as such:

25 = (100)(1/2)^(t/1000) --> now you divide by 100 on both sides to simplify the expression. YOU CANNOT DIVIDE OUT THE 1/2 BECAUSE IT HAS AN EXPONENT ATTACHED TO IT!

(25/100) = (1/2)^(t/1000) --> next, if you are firmilliar with solving for exponents using logs, you write log on both sides in front of each side, then bring the exponent down in front of the log ON THE SIDE WITH THE VARIABLE.

log(25/100) = (t/1000)log(1/2) --> now you can divide by log(1/2) on both sides to simplify the equation:

((log(25/100))/(log(1/2)) = (t/1000)
2 = (t/1000) --> now you can multiply by 1000 on both sides to isolate "t".

2000 = t --> That's it! I hope this was of help! :)