SOLUTION: Hi!!!!! I tried attempting this really hard problem on a homework assignment and I have no clue what i'm doing wrong, it's a half life question and i keep getting the wrong sort of

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hi!!!!! I tried attempting this really hard problem on a homework assignment and I have no clue what i'm doing wrong, it's a half life question and i keep getting the wrong sort of      Log On


   



Question 925599: Hi!!!!! I tried attempting this really hard problem on a homework assignment and I have no clue what i'm doing wrong, it's a half life question and i keep getting the wrong sort of answer, I hope you're ready :D "An accident in 1986 at Cherynobyl play in the Ulraine released a large amount of plutonium (Pu-239) into the atmosphere. The half-life of Pu-239 is about 24,110 years.
a) Find the decay constant
b) Use the function N(t) = N0e^-kt to find what remains of an initial 20 grams of Pu-239 after 5000 years.
c) How long will it take for these 20 grams to decay to know gram? Go down to the Years, Days, Hours, Minutes, and Seconds.. I got A! so that won't be a problem, maybe you could try and solve it too? I got a weird number for b - 17.3? Is that odd? I mean... isn't it supposed to go down? And what's the nonsese with c? Any help at all would be great, thanks sO much!

Found 3 solutions by josgarithmetic, MathTherapy, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
N%28t%29=s%2Ae%5E%28-kt%29
N(t), amount in grams at time t
s, initial amount (at time 0, when the plutonium was released?); think "s"tart;
e, the natural logarithm base
k, a constant
t, time passage in years

The half-life fact allows you to find the value for k.
In general, based on the decay equation
1%2F2=1%2Ae%5E%28-k%2A24110%29, although this is really for Plutonium-239.
ln%281%2F2%29=ln%281%2Ae%5E%28-k%2A24100%29%29
ln%281%2F2%29=ln%281%29%2Bln%28e%5E%28-k%2A24100%29%29
ln%281%2F2%29=0%2B%28-kt%29%2Aln%28e%29
ln%281%2F2%29=-kt
-ln%282%29=-kt
kt=ln%282%29, combining two steps; and recall that for here, half-life is...
k...
highlight_green%28k=ln%282%29%2F24110%29
OR
highlight_green%28k=2.8749%2A10%5E%28-5%29%29

You should have no trouble with question part (b), because now you have the function highlight%28N%28t%29=s%2Ae%5E%28%28-2.8749%2A10%5E%28-5%29%2At%29%29%29; and you are given s=20, t=5000.

The way to handle question part (c) is to solve the decay model equation function for time, t.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Hi!!!!! I tried attempting this really hard problem on a homework assignment and I have no clue what i'm doing wrong, it's a half life question and i keep getting the wrong sort of answer, I hope you're ready :D "An accident in 1986 at Cherynobyl play in the Ulraine released a large amount of plutonium (Pu-239) into the atmosphere. The half-life of Pu-239 is about 24,110 years.
a) Find the decay constant
b) Use the function N(t) = N0e^-kt to find what remains of an initial 20 grams of Pu-239 after 5000 years.
c) How long will it take for these 20 grams to decay to know gram? Go down to the Years, Days, Hours, Minutes, and Seconds.. I got A! so that won't be a problem, maybe you could try and solve it too? I got a weird number for b - 17.3? Is that odd? I mean... isn't it supposed to go down? And what's the nonsese with c? Any help at all would be great, thanks sO much!
You did get the correct amount for b), but with the wrong sign. It should be 17.3 gm, and NOT - 17.3 gm.
The reason why you got a - 17.3 is because you used the WRONG formula. The GROWTH/DECAY formula is:
Now, if k < 0, then that signifies DECAY/DECREASE, and k being > 0 signifies GROWTH!
For c), does "know gram" mean NO GRAMS, as in ZERO (0) grams?

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


a) find the decay constant

Use the general equation and the fact that a sample decays to half its original in 24110 years.

1%2F2+=+e%5E%28-24110k%29
ln%28.5%29+=+-24110k
k+=+ln%28.5%29%2F-24110+=+-.00002875 to 4 significant figures

b) find how much remains of 20g after 5000 years

Use the formula with the value of k you found.

20%2Ae%5E%28-.00002875%2A5000%29+=+17.322

The statement of part c) makes no sense grammatically, so there is no way to know what the intended question was....