Question

A certain radioactive isotope has a half-life of 11 days. If one is to make a table showing the half-life decay of a sample of this isotope from 32 grams to 1 gram; list the time (in days, starting with t = 0) in the first column and the mass remaining (in grams) in the second column, which type of sequence is used in the first column and which type of sequence is used in the second column?

Answer 1

@perl :)

Answer 2

do you know the formula for half-life?

Answer 3

no :/ this lesson told us basically nothing

Answer 4

exponential functions?

Answer 5

I am somewhat familiar with exponential functions.

Answer 6

good

Answer 7

so at let us make sure we agree that \(t=0 \rightarrow t=11 \) half of that species (whatever radioactive isotope) will be gone

Answer 8

yes, I would agree with that

Answer 9

next, is that we need to identify the constant in which the decay occurs by using \(0.5 = e\large^{11k} \)

Answer 10

understand that we are calculating by "days" from the basis of the problem presented

Answer 11

would you try to solve for k?

Answer 12

time = half-life * log (begng amount / ending amt) / log 2

Answer 13

let us do it this way then I will teach you how to derive for any rate-related problem later without memorizing anything

Answer 14

I honestly am completely clueless on how to solve for k.

Answer 15

ah, so this is the part of the exponential functions that you will have to make use of logarithmic functions

Answer 16

have you got any idea about logs?

Answer 17

my formula is easier to use but you want to use that other one huh?
'k' (sometimes called lambda) is equal to natural log (2) / half-life

Answer 18

everyone's formula is always easier for the lazy mind, learning how that formula arose is for the inquiring mind

Answer 19

Yes, I know logarithms some, I am unable to do anything on the computer on OS. It just completely froze.

Answer 20

okay inquirer
k = 0.69314718056 / 11

Answer 21

or k = 0.0630133801

Answer 22

we can solve for k by using \(ln(0.5) = 11k \) then manipulate so you have the equation as k

Answer 23

the reason we are not trying to get the actual value is so that we do not fall into the traps of "estimation." we will do this in the very end

Answer 24

I am VERY familiar with half life calculations
here is a calculator I wrote
http://www.1728.org/halflife.htm
it does not do anything with estimation

Answer 25

-0.693147 = 11k
-0.063013\(\overline{36}\) = k

Answer 26

would it suffice and would it hurt anyone's ego for now if we just used \(\huge \frac{ln(0.5)}{11} = k \)

Answer 27

that would be fine with me

Answer 28

that works too :)

Answer 29

next, pay attention to the instruction:
If one is to make a table showing the half-life decay of a sample of this isotope from 32 grams to 1 gram; list the time (in days, starting with t = 0) in the first column and the mass remaining (in grams) in the second column

Answer 30

got a paper and pen or a word or even spreadsheet software so it is neat-looking

Answer 31

Okay sorry, I put the columns in the wrong order

Answer 32

ye, paying to the instruction is crucial specially in the fields of science like chemistry and physics

Answer 33

This is pretty complicated for me, but I understand it better now :)

Answer 34

haha
so we will try to go down from day 0 then increment of 1 until we get really near to the value of the mass to 1 gram so we don't have to waste so much lines

Answer 35

=) thank you both

Answer 36

so you know how to do the rest?

Answer 37

I mean we are not done

Answer 38

time(days) remaining amt
0 32
0.5038406 31
1.0242034 30
1.5622091 29
2.1190959 28
2.6962375 27
3.2951631 26
3.9175819 25
4.5654125 24
5.2408185 23
5.9462522 22
6.6845083 21
7.4587910 20
8.2727974 19
9.1308250 18
10.0379087 17
11.0000000 16
12.0242034 15
13.1190959 14
14.2951631 13
15.5654125 12
16.9462522 11
18.4587910 10
20.1308250 9
22.0000000 8
24.1190959 7
26.5654125 6
29.4587910 5
33.0000000 4
37.5654125 3
44.0000000 2
55.0000000 1

Answer 39

there it is with the columns in the correct order AND I deleted the previous answer

Answer 40

or you can start from 0th day then jump to 11th day since we already know that half of the mass will remain cutting our work more systematically