Question

The nuclear waste from an atomic energy plant decays exponentially at a rate of 3% each century.
(a) Which fraction of the original amount remains after one century? Two centuries?
(b) What is the base of the exponential function involved?
(c) If 150 pounds of nuclear waste is disposed of, how much of it will still remain after 10 centuries?

Should I use the half life formula, if so how with only the info given, I know the decay should be something in the form of 1.03 to represent 3% but not sure how to proceed.

Answer 1

Are you familiar with exponential decay?

Answer 2

\[N(t) = N_0e^{kt}\]

Answer 3

Yes, I have read up on it and was wanting to use it but wasn't sure how to approach it, I was thinking k would = 1.03, but without other information was lost on what to do.

Answer 4

I know N(1) can represent say 1 century so forth, but I don't even know 3% of what amount is decaying.

Answer 5

And sorry, I got the original expression incorrect it should be raised to the -kt.
So what you're looking for in part a is simply:
\[\frac{N(t)}{N_0}\]

Answer 6

So then for part A, you simply have:
\[e^{-kt}\]
where "k" is the rate. 3% = (0.03), and t = time in centuries. So solve it for t = 1 and t = 2.

Answer 7

So I'd simply do \[e ^{-0.03x1} \] and\[e ^{-0.03x2}\] ?

Answer 8

You got it.

Answer 9

Since we've solved for the ratio, that is all that is left. You'll get a percentage.

Answer 10

Does that mean, the answers just given are where I'd stop?

Answer 11

What did you get....?

Answer 12

1.0304545.... Repeater on 45

Answer 13

That...would be incorrect. What did you do?

Answer 14

Ahh I forgot the negative. With the neg got: 0.97044553...

Answer 15

Remember, e raised to the NEGATIVE power....you forgot the negative.

Answer 16

Correct. So, after 1 century, only about 97% of the original amount remains. After 2 centuries, only about 94% of the original amount remains.

Answer 17

So for part B, what do you think the base is?

Answer 18

Would that simply mean -0.03xD^n

Answer 19

For part B??

Answer 20

yes, I am confused by what they mean as the "base" of the etc..

Answer 21

Okay, well look at the exponential term "-kt"....what is the base?

Answer 22

-k, as the t is the variable?

Answer 23

No, you're looking at the BASE. So if you had 2^4, for example. What is the base?

Answer 24

base of 2^4, 2?

Answer 25

That's right, now apply that to e^(-kt), what is the base? (Forget the fact that k and t are multiplied, this is irrelevant....k is just a constant)

Answer 26

base e?

Answer 27

Thats right.

Answer 28

excellent, I think I was over thinking those terms. Would it ultimately mean the smallest intial starting point, is the "base" where by the variables are divided into?

Answer 29

Well, it simply means this, if x is the variable then:
\[a^x\]
a = base
x = exponent.

Forget the fact that x (or t in this case) is multiplied by some constant. The fact is that the base is a, the exponent is x with some constant.

Answer 30

So for part C, what have you?

Answer 31

For C, I am thinking e^-0.03x10, to get the decay after 10 centuries first up. But I am not sure where to use the 150 pounds, or essentially 150.

Answer 32

That is going to be your initial (or N0) term. You want to see how much remains as a function of time N(t).

Answer 33

So N(t) is the unknown.

Answer 34

N(10)=N150^e^-0.03x10 ?

Answer 35

You're not raising e to any power:
\[N(t) = 150 e^{(-0.03)(10)}\]

Answer 36

Sorry, you're not raising the 150 to any power. You're simply raising e to the (-0.03)(10) power. Then multiplying that result to 150. You see, the e^(-kt) term is going to be some fraction....you're multiplying 150 by that fraction to give you how much REMAINS.

Answer 37

ahh yes, ignore that ^ after 150, I was trying to imitate syntax, but yeah 150xe^etc was what I meant. So it would be what you wrote but \[N(10)=150e ^{(-0.03)(10)}\]

Answer 38

So what have you?

Answer 39

1455.6683 after computing that equation.

Answer 40

Always ask yourself "Does this make sense?" after you solve these. Does it make sense for a DECAY equation to give you *MORE* weight than you started with??

That is not what I got. Evaluate:
\[e^{(-0.03)(10)}\]
What is the result?

Answer 41

I know what you did ;) You need to pay careful attention. You're raising e to the (-0.03 multiplied by 10) power. *NOT* e to the -0.03 power THEN multiply by 10. Mind your parenthesis.

Answer 42

for e^(-0.03)(10) = 9.70445533,

I did 150x e^etc earlier for that other result.

Answer 43

-0.03 * 10 = -0.3, then e^-0.3 = 0.74

Answer 44

e^-0.3 was much better lol :)

Answer 45

I got 0.74...etc

Answer 46

so do I now proceed by doing 150*0.74, thus 74% of 150 will remain after 10 centuries?

Answer 47

Thats right. So roughly 111lb will remain after 1000 years. Long time, eh?

Answer 48

hehe yep, 111 pounds left after a millennia. Thank you so much for your help, I sometimes just need guidance to understand the intricate details for solving Math. You have been a big help.

Answer 49

No problem. Remember, always ask yourself if it makes sense. Math isn't just numbers. The equations tell a story.