What’s your issue with radiometric dating

?

Radiometric dating gets more and more accurate as we get better at it

Yes it can

It’s dated rocks to 1% margin of error

There will always be error bars

But the point is is it’s extremely precise

Study a little how we use radiometric dating

Do you even know the equation we use for Potassium Argon then?

What is it?

Then you haven’t studied it

That’s

That’s what we use to date things

A..

Are you serious?

Listen

I’m going to attempt to give you a lesson in radiometric dating

And I’m gonna hope that by the end of this, you throw out the idea that it’s inaccurate

We have an equation

N(t) = N_0 e^-kt

Where N_0 = the initial value

e is Eulers number

k is the constant of proportionality

and t is time

Do you understand?

Okay

We know that argon has 8 electrons, and thus has little to no reactivity

Do you agree

Okay

We know that by using quantum physics, we can calculate the half life of potassium 40, an isotope of potassium, to be 1.25 billion years

Do you agree?

Yes

We know that K-40, potassium 40, decays into two atoms

since it’s radioactive

Argon 40, and Calcium 39

We know that 11% is argon 40, and 89% is Ca 39

No

One of the neutrons turns into an electron

Pretty sure

Maybe proton

And if we have some sort of volcanic eruption, the lava is illiquid

Liquid**

That means that any argon 40 in there, since it’s low reactivity, is free to move out of the lava

It’s probably an error

@PhumoZTYPE anyway

Because of this, we know that any argon 40 in the rock must have came from the parent element

Which is K-40

Good

Now, we can get to the fun part

It would be easier if I could send pictures

We can take our original equation, N_0 times e^-kt = N(t)

I’ll have to read up on that then

It’s irrelevant to this discussion though

All we care about is Argon 40

@PhumoZTYPE

We can plug in 1.25B to our equation for t

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

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

N(t) is the amount of K-40

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

Why?

No no

We don’t even need to measure it

We can actually calculate the initial amount

I’ll show you

Bare with me

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

Which is half the initial amount

We can divide N_0 on both sides

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

Apply ln to both sides

Or the natural logarithm

This’ll get you ln(1/2)= -k*1.25B

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

Are you still with me?

Now

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

And our current A-40 to be 0.01 mg

This is something we can find using normal means

We want to know “how old is this rock”

We have this equation
1mg (current amount) = N_0*e^-kt

And we already know k

All we need is the initial amount

We have 1mg, our current, plus .01/11%

Because remember, we have a certain amount of K-40 that had to decay into .01mg of Argon

And that argon is 11% of that

So .01/11% will get us the amount that decayed

Still with me?

Now

Doing some fancy maths

We can isolate t

And calculate it to be 157 million years

This

Thus

We know the age of our rock

The rock as a whole

Do you see why this works now?

What don’t you understand

Wow

We didn’t **assume** 157 million years