Question

The half life for the radioactive decay of potassium-40 to argon-40 is 1.26 X 10^9 years.

Suppose nuclear chemical analysis shows that there is 0.289 mmol of argon- 40for every 1.000 mmol of potassium-40in a certain sample of rock. Calculate the age of the rock.
Round your answer to 2 significant digits.

Answer 1

Have you tried solving it? Haw far have you got? (Use the exponential decay formula)

Answer 2

I get stuck on these word problems mostly becuase it does not speak english. It all sounds like a forgien language to me.

Answer 3

Is the formula A = 28(1-0.5)^42?

Answer 4

The exponential decay formula is y = ae^(rt), where 'y' is the remaining amount of radioactive substance after time 't' and 'a' is the original amount of radioactive substance.

Half life is given 1.26 x 10^9 years. So when t = 1.26 x 10^9, the remaining amount 'y' is half of the original amount 'a'. Hence put y = a/2 and t = 1.26 x 10^9 in the formula to get

a/2 = ae^(r * 1.26 * 10^9)

Can you try to solve the equation above for 'r' ?

Answer 5

I have to put r on the other side of the equation in order to solve right?

Answer 6

Yes, and in order to do that, you need to first remove 'a' by dividing the equation by 'a' on both sides

Answer 7

Wouldn't that just make a on both sides disappear or does it put 1 in place of a?

Answer 8

Yes, it would, and it is the only logical step since making 'a' disappear from the equation will have you left with only one unknown, r (e is the natural number, not a variable)

Answer 9

How is e a number?

Answer 10

e is the symbol for a mathematical constant just like pi

Answer 11

Oh, so /2=e^(r*1.26* 10^9)

Answer 12

Good!

Any idea how we would solve further?

Answer 13

Put 2.718281828 as the number for e?

Answer 14

No you can take natural log on both sides to make e disappear

Answer 15

what does natural log mean?

Answer 16

Oh, do I divide both sides by e?

Answer 17

no, log is a function which you can consider the opposite of 'e^()', so taking it on both sides of the equation you get

log(0.5) = log(e^(r*1.26*10^9))

Answer 18

value of log(0.5) is -0.693

Answer 19

Use this rule now: log(e^(anything)) = (anything)*log(e),
so you can simplify the right hand side to r*1.26*10^9 log(e)
Now put log(e) = 1. So you get the equation

-0.693 = r*1.26*10^9

Answer 20

now can you solve this equation for 'r' ? This will give you the value of rate of change and hence will complete half of the question's solution

Answer 21

so I again divide both sides by r wich will give us, r-0.693 =1.26*10^9

Answer 22

to get the value of 'r', remove the numbers along with it. so you would divide by 1.26 * 10^9 on both sides, not r

Answer 23

Oh, Ok so \[\frac{ -0.693 }{ 1.26*10^{9} }=r\]

Answer 24

Good job! Need to simplify that..

Answer 25

so I need to put the signintific notation in to regular numbers right?

Answer 26

don't do anything with the 10^9 except bring it on the numerator by making the 9 negative,. otherwise just divide the other numbers

Answer 27

I get -0.55

Answer 28

Good thats what I got. Now we have the value of rate r = -0.55

Answer 29

it is r= -0.55 * 10^-9

Answer 30

Now we come to the scond part of the question. It says that for every 1 mol potassium - 40 you have 0.289 mol of argon.

Can you guess what might be the total amount of potassium - 40 in the sample when there was no argon formed at the beginning?

Answer 31

19-40 is 21

Answer 32

oh, -21

Answer 33

How did you get that 21? It is not given in the question

Answer 34

Since 0.289 moles of argon are there it means there were as many moles of potassium at the beginning in addition to the 1 mole already present in the sample

Answer 35

19 is the atomic number of potassium. I thought you were saying potassium (K) - 40.

Answer 36

so 1.289

Answer 37

Good!

Answer 38

So we have initial amount of potassium 1.289, remaining amount of potassium 1.0 and rate -0.55 * 10^-9. Plug these in the equation Y = Ae^(rt) to get the value of t

Answer 39

rt is rate I know that but is Ae the potassium or the Y?

Answer 40

We are solving for Y right?

Answer 41

r is the rate , t is the time required to convert 'a' moles of potassium into 'y' moles

Answer 42

So Y = 1.289e^(-0.55*10^-9t)
did I do that right?

Answer 43

Right! and put y = 1.0 as well

Answer 44

1.0 = 1.289e^(-0.55*10^-9t) and we are solving for "t" time

Answer 45

yes again use the log procedure we did for the first equation to get the value of t

Answer 46

log(1.0) = 1.289 log(e^(-.55*10^-9t))

Answer 47

You can't just keep 1.289 out of the log just because it is a number. That does not occur with logs. Instead, to simplify things, divide by 1.289 on both sides so that you end up with a log(1/1.289) on the left hand side

Answer 48

I get 461575861.6 which is 4.6*10^-8

Answer 49

Excellent so you got the answer! :)

Answer 50

Really! Woohoo!

Answer 51

So I have one more question along these same prameters.

Answer 52

Suppose the amount of a certain radioactive substance in a sample decays from 3.80 mg to 2.50 mg over a period of 49.5 years. Calculate the half life of the substance.

Answer 53

Okay its the same as the previous one. First find the value of 'r'

Answer 54

btw can you start a new thread please because this one is becoming too long to scroll!

Answer 55

Ok, I posted it in another question.

Answer 56

@Mymathhomework I'm on the new question now