Message from @PhumoZTYPE

Discord ID: 603634554846314506


2019-07-24 17:05:21 UTC  

K-40 to Ar-40 is by electron capture

2019-07-24 17:05:35 UTC  

It’s irrelevant to this discussion though

2019-07-24 17:05:44 UTC  

All we care about is Argon 40

2019-07-24 17:05:49 UTC  

I shall keep my mouth shut then

2019-07-24 17:05:56 UTC  
2019-07-24 17:06:02 UTC  

Yeah

2019-07-24 17:06:07 UTC  

I’m listening

2019-07-24 17:06:09 UTC  

We can plug in 1.25B to our equation for t

2019-07-24 17:06:15 UTC  

And set it equal to 1/2 N(t)

2019-07-24 17:06:32 UTC  

Why lol

2019-07-24 17:06:46 UTC  

Because in 1.25 billion years, we’ll have half the amount of K-40

2019-07-24 17:06:59 UTC  

Oh yess

2019-07-24 17:07:00 UTC  

N(t) is the amount of K-40

2019-07-24 17:07:18 UTC  

So we have 1/2 N_0=N_0 e^-k*1.25B

2019-07-24 17:07:36 UTC  

But isn’t that based on the assumption that potassium 40 is like that

2019-07-24 17:07:47 UTC  

Why?

2019-07-24 17:07:55 UTC  

It has been measured

2019-07-24 17:07:58 UTC  

No no

2019-07-24 17:08:01 UTC  

We don’t even need to measure it

2019-07-24 17:08:08 UTC  

We can actually calculate the initial amount

2019-07-24 17:08:12 UTC  

How

2019-07-24 17:08:16 UTC  

I’ll show you

2019-07-24 17:08:20 UTC  

Bare with me

2019-07-24 17:08:39 UTC  

I’m trying yeah

2019-07-24 17:08:41 UTC  

We have So we have 1/2 N_0=N_0 e^-k*1.25B

2019-07-24 17:08:48 UTC  

Which is half the initial amount

2019-07-24 17:08:58 UTC  

We can divide N_0 on both sides

2019-07-24 17:09:10 UTC  

Yeah

2019-07-24 17:09:12 UTC  

To get 1/2 = e^-k*1.25B

2019-07-24 17:09:19 UTC  

Apply ln to both sides

2019-07-24 17:09:28 UTC  

Or the natural logarithm

2019-07-24 17:09:48 UTC  

This’ll get you ln(1/2)= -k*1.25B

2019-07-24 17:10:11 UTC  

Divide -1.25 B and you get k= ln(1/2) / -1.25B

2019-07-24 17:10:37 UTC  

Are you still with me?

2019-07-24 17:10:50 UTC  

Yup

2019-07-24 17:10:59 UTC  

Now

2019-07-24 17:11:12 UTC  

Let’s say we find our current amount of K-40 to be 1mg

2019-07-24 17:11:22 UTC  

And our current A-40 to be 0.01 mg

2019-07-24 17:11:48 UTC  

This is something we can find using normal means

2019-07-24 17:12:06 UTC  

We want to know “how old is this rock”

2019-07-24 17:12:29 UTC  

Yeah