a certain radioactive isotope has a half-life of 24,000 years. If there are 4 grams of the isotope remaining after 21,500 years, how many grams were there originally????
2006-11-19 11:18:36 · 6 answers · asked by sportsrlife 1 in Science & Mathematics ➔ Astronomy & Space
"Give a student an answer, and their grade on that assignment will go up a point, teach them how to find their answer and their score on the final exam will improve!"
Radioactive decay is an exponential thing; for every 24000 years that elapses in this problem, half the remaining material vanishes. If you express this as an algebraic equation you can figure this stuff out easily. The hard part is figuring out the equation.
Lets say you start out with X_0 grams of material. The amount of material remaining is X:
X= X_0*(1/2)^(t/24000)
Where t is given in years. See how this works? if t = 0, then X=X_0*1/2^0 power. Anything raised to the 0 is always 1, so X=X_0. At t=24000 years, X=X_0*1/2^(1), so X=X_0*1/2; half as much is left.
In this problem you are given t, and X and are asked to find X_0 so you use a calculator to turn 1/2^(21500/24000) into just a number and then you have
4 grams=X_0*number
Pretty easy to solve that for X_0. The point is, once you write down the equation that describes the system, you can manipulate it to solve for any of the terms in it, as long as you know the other two terms.
Incidentally, I get a different answer than those other two folks who agree with one another. They are wrong, but not by a lot. Work it out for yourself.
[Edit: In fact I get the same result IMI does, so my method does work. Her answer was not posted when I started working on mine.]
2006-11-19 12:26:04 · answer #1 · answered by Mr. Quark 5 · 1⤊ 0⤋
I got something different
N(t) = N(0)*2^((-t)/(halflife))
Here we are given :
N(t) (mass at time t) = 4grams
t=21500 and
halflife=24000
We want to solve for N(0) the original mass
4=N(0) * 2^ (-21500/24000)
simplify the exponent
4=N(0) * 2^ (-.895833)
take two to the exponent
4=N(0) * .537437
divide each side by .537...
7.44274~N(0)
There were originally about 7.4 grams
2006-11-19 12:10:27 · answer #2 · answered by iMi 4 · 0⤊ 0⤋
1500 yrs * 55% = do the math
2016-05-22 04:54:12 · answer #3 · answered by Anonymous · 0⤊ 0⤋
7.1666666666666666666666666666667 grams
2006-11-19 11:22:55 · answer #4 · answered by Walking Man 6 · 0⤊ 1⤋
ther are 13000
2006-11-19 11:21:29 · answer #5 · answered by Patrick L 2 · 0⤊ 1⤋
7.1666.. gms
2006-11-19 11:27:06 · answer #6 · answered by ramshi 4 · 0⤊ 1⤋