Age of Meteorite: Calculating Half-Life

[Rectifier]

The problem statement
The ratio between stable Argon atoms($^{40}Ar$) and radioactive Potassium atoms($^{40}K$)in a meteorite is 10.3. Assume that these Ar atoms are produced by decay of Potassium-atoms, whose half-life is $1.25 \cdot 10^9$ years. How old is the meteorite?
Translated from Swedish.

The attempt at a solution
Radiactive decay can be calculated with $N(t) = N_0e^{- \lambda t}$ where t is time and lambda is constant. Half-life gives us $\frac{1}{2} = N_0e^{- \lambda t}$

And we know that
$\frac{N_{r}}{N_s}=10.3$ and $N_{r} + N_s = N_0$ thus $N_r = \frac{N_0}{11.3}$
$N(t) = N_0e^{- \lambda t}$ and $N_r = \frac{N_0}{11.3}$ give us $\frac{N_0}{11.3} = N_0e^{- \lambda t}$ and $t=4.3 \cdot 10^9$ which is wrong :/.

Can someone help please?

 

[DrClaude]

The problem statement
The ratio between stable Argon atoms($^{40}Ar$) and radioactive Potassium atoms($^{40}K$)in a meteorite is 10.3.

$\frac{N_{r}}{N_s}=10.3$

The way I read it, there is 10.3 times more argon than potassium, not the other way around.

 

[Rectifier]

The way I read it, there is 10.3 times more argon than potassium, not the other way around.

Tanks for the reply!

So it should be:
$\frac{N_{s}}{N_r}=10.3$ and $N_{s} + N_r = N_0$ thus $N_r = \frac{N_0}{11.3}$
$N(t) = N_0e^{- \lambda t}$ and $N_r = \frac{N_0}{11.3}$ give us $\frac{N_0}{11.3} = N_0e^{- \lambda t}$ and $t=4.3 \cdot 10^9$

and I get the same answer :D

 

[DrClaude]

Tanks for the reply!

So it should be:
$\frac{N_{s}}{N_r}=10.3$ and $N_{s} + N_r = N_0$ thus $N_r = \frac{N_0}{11.3}$
$N(t) = N_0e^{- \lambda t}$ and $N_r = \frac{N_0}{11.3}$ give us $\frac{N_0}{11.3} = N_0e^{- \lambda t}$ and $t=4.3 \cdot 10^9$

and I get the same answer :D

I didn't notice that you had made a second error cancelling the first one!

I notice now that

Half-life gives us $\frac{1}{2} = N_0e^{- \lambda t}$

is not correct.

What is the value of $\lambda$?

 

[Rectifier]

I didn't notice that you had made a second error cancelling the first one!

I notice now that

is not correct.

What is the value of $\lambda$?

Is it $\lambda = \frac{ln2}{T_{1/2}}$ ?

EDIT:
Thus $\lambda = \frac{ln2}{T_{1/2}} =\frac{ln2}{1.25 \cdot 10^9 }$

 

[DrClaude]

Is it $\lambda = \frac{ln2}{T_{1/2}}$ ?

EDIT:
Thus $\lambda = \frac{ln2}{T_{1/2}} =\frac{ln2}{1.25 \cdot 10^9 }$

That's correct.

I made the calculation myself, and find the same value as you. Why do say the answer is wrong?

 

[Rectifier]

That's correct.

I made the calculation myself, and find the same value as you. Why do say the answer is wrong?

My teacher told me so. Well I guess that he is wrong then.

 

[DrClaude]

My teacher told me so. Well I guess that he is wrong then.

Didn't he give any hint as to what was wrong?

The only thing I can see is that your answer does not have the correct number of significant digits.

 

[Rectifier]

Didn't he give any hint as to what was wrong?

The only thing I can see is that your answer does not have the correct number of significant digits.

Ah sh*t :D
"Svara med två decimaler"

Then its 4.37 :D

 

[DrClaude]

Ah sh*t :D
"Svara med två decimaler"

Then its 4.37 :D

Precis!

You could also infer it from the fact that both the ratio and the half-life are given with three significant digits.

 

[Rectifier]

Precis!

You could also infer it from the fact that both the ratio and the half-life are given with three significant digits.

Thank you for your help!