Question

The decay of uranium is modeled by D=D1 x 2^-kt. If it takes 6 years for the mass of uranium to halve, find the percentage remaining after 2 years

Answer 1

"k" in the equation is called the decay constant of the sample i.e., reciprocal the time taken for half the initial amount (no matter how much iniial, but half of it) to decay.
k = 1 / (half-life time)

the question asks you to find
% remaining = (initial amount - lost) *100 / (initial amount)
= (D1-D)*100/D1
=(1 - D/D1)*100
given t=2

Answer 2

I didn't get what you were trying to explain

Answer 3

decay const :
k = 1 / (time to decay to half) = ?

Answer 4

I'll write it down on paper to see if it makes sense

Answer 5

\[k=\frac{1}{t_{1/2}}\]
\(t_{1/2}\) is given in the problem

Answer 6

I still can't get it

Answer 7

I mean the whole formula you mentioned above

Answer 8

you are given\[D=D_1e^{-kt}\]
correct?

Answer 9

How to find the percentage and how can I find D and D1

Answer 10

Ok

Answer 11

It's 2 instead of e

Answer 12

D1 = total amount initially present
D = amount left after time t

Answer 13

correct, "2" instead of "e"

Answer 14

agree?
the question is asking you
what % is left after time t

Answer 15

Now I'm beginning to understand

Answer 16

so, they are asking for \({D\over D_1}\times100\)

Answer 17

Right

Answer 18

so,\[\%left = 2^{-kt},\qquad k={1\over t_{1/2}}\]
\(t_{1/2}\) is the time taken to reduce to half the amount

Answer 19

Yep

Answer 20

And we have the value of t which is 2

Answer 21

you may proceed from here.
1. find k using given \(t_{1/2}\) (using the equation on the right)
2. use that k and given "t" (using equation to the left)

that is your answer

Answer 22

yep

Answer 23

Would k =1 ?

Answer 24

no, step "1"

Answer 25

Sorry 1/2?

Answer 26

what is \(t_{1/2}\)?

Answer 27

Isn't it 2?

Answer 28

how? it is provided in the problem

Answer 29

"6"

Answer 30

what happens in "6" years?

Answer 31

The uranium halves

Answer 32

that is \(t_{1/2}\) =half life time

Answer 33

So k is 1/6?

Answer 34

half life = time taken for the amount of material to reduce to half-its initial amount

Answer 35

yes k=1/6

Answer 36

:) sorry man I'm giving you a hard time

Answer 37

thats kool.

Answer 38

Now to find the % I have to insert k (1/6) and t (2) into the equation?

Answer 39

that's it!

Answer 40

Then I multiply by 100

Answer 41

yep

Answer 42

Could you please show me how just to make sure

Answer 43

\[\%=2^{-{1\over6}\times2}\times100=2^{-1\over3}\times100=?\]

Answer 44

79%

Answer 45

I've got one more thing to ask you about

Answer 46

yes.. sounds reasonable.

Answer 47

You wrote previously above that % = (D1-D) x 100/D1

Answer 48

Where did you deduct it from?

Answer 49

yes.. my bad.. I thought D represented amount "lost"
but D represents amount "left"

Answer 50

Is that equation still correct ?

Answer 51

the question does not tell us about this abiguity

Answer 52

Is that another way of finding % if we were given D and D1?

Answer 53

those are the two ways.

Answer 54

based on the interpretation of the words of the question

Answer 55

Is there any other way other than these two ways?

Answer 56

Anyways thanks a lot for your help. I really appreciate it.

Answer 57

you are welcome