Determine the value of a decay constant

[Kingyou123]

Homework Statement

What's up guys, I'm stuck on #2 of this worksheet,http://luc.edu/faculty/dslavsk/courses/phys111/homework/phys111-2015hw1.pdf
The half - life of C - 14 is 5730 years. Determine the value of its decay constant (and express the answer in appropriate MKS units).

2.The attempt at a solution
I'm not sure if I'm doing this correctly , but I set up my formula like this , 5730=14e-(constant of decay)5730

 

[SammyS]

Homework Statement

What's up guys, I'm stuck on #2 of this worksheet,http://luc.edu/faculty/dslavsk/courses/phys111/homework/phys111-2015hw1.pdf

2.The attempt at a solution
I'm not sure if I'm doing this correctly , but I set up my formula like this , 5730=14e-(constant of decay)5730

Hello . Welcome to PF !

You will get a better response if you display the question so that it can be read directly. The following is what I was able to obtain from your pdf file using Window's "Snipping Tool" & easily display here.

Better yet, you could type out the problem. I makes it easier to respond to, and it shows initiative on your part.

As far as your attempt at a solution goes, it looks like you simply picked out some numbers at random from the problem statement and then plugged them into the given equation.

 

[Kingyou123]

Hello . Welcome to PF !

You will get a better response if you display the question so that it can be read directly. The following is what I was able to obtain from your pdf file using Window's "Snipping Tool" & easily display here.
View attachment 87850

Better yet, you could type out the problem. I makes it easier to respond to, and it shows initiative on your part.

As far as your attempt at a solution goes, it looks like you simply picked out some numbers at random from the problem statement and then plugged them into the given equation.

I've been using this fourm. https://www.physicsforums.com/threads/decay-constant.24169/ , as a guide and I can't figure out what to do... it's the first assignment of the semester and we haven't gone over this...

 

[SammyS]

I've been using this forum. https://www.physicsforums.com/threads/decay-constant.24169/ , as a guide and I can't figure out what to do... it's the first assignment of the semester and we haven't gone over this...

C-14 is carbon 14. That's just an identifier for the type of nucleus being addressed. The number 14, will not be involved anywhere in the solution.

The problem statement gives many hints, maybe even telling you much of what you need to do to solve the problem.

If you start with N₀ C-14 nuclei, then after 5730 years (one half-life), how many C-14 nuclei remain in your sample?

That quantity is what you put in for N(5730) .

 

[Kingyou123]

C-14 is carbon 14. That's just an identifier for the type of nucleus being addressed. The number 14, will not be involved anywhere in the solution.

The problem statement gives many hints, maybe even telling you much of what you need to do to solve the problem.

If you start with N₀ C-14 nuclei, then after 5730 years (one half life), how many C-14 nuclei remain in your sample?

That quantity is what you put in for N(5730) .

Omg, you just made it so easy... so the formula would be 5370=(.5)e^(-(decay)5370) or would I set the entire thing to zero?

 

[SammyS]

Omg, you just made it so easy... so the formula would be 5370=(.5)e^(-(decay)5370) or would I set the entire thing to zero?

No.

And you didn't answer the main question I asked..

If you start with N₀ nuclei, then how many will remain after one half-life has elapsed?

 

[Kingyou123]

No.

And you didn't answer the main question I asked..

If you start with N₀ nuclei, then how many will remain after one half-life has elapsed?

Half of the N0 nuclei would remain.

 

[SammyS]

Half of the N0 nuclei would remain.

Correct.

That's what goes on the left hand side of the equation.

 

[Kingyou123]

Correct.

That's what goes on the left hand side of the equation.

So the formula would start looking like this .5=(.5)e^-(decay)5730

 

[SammyS]

So the formula would start looking like this .5=(.5)e^-(decay)5730

How did you get that?

Where does the .5 come from on both sides?

Isn't the left side N₀/2 , and the right side N₀ times the exponential ?

 

[Kingyou123]

How did you get that?

Where does the .5 come from on both sides?

Isn't the left side N₀/2 , and the right side N₀ times the exponential ?

I assumed that .5 is half, I get confused when I don't enter values in. So No/2=N0e^-(decay)5730? The thing that confuses me is what values to put in the fourmla that I have now.

 

[SammyS]

I assumed that .5 is half, I get confused when I don't enter values in. So No/2=N0e^-(decay)5730? The thing that confuses me is what values to put in the fourmla that I have now.

You should use another set of parentheses to make that technically correct.

0.5⋅N₀=N₀e^(-(decay)5730) .

Just solve for the decay constant. First step is to divide both sides by N₀ . Right ?

 

[Kingyou123]

You should use another set of parentheses to make that technically correct.

0.5⋅N₀=N₀e^(-(decay)5730) .

Just solve for the decay constant. First step is to divide both sides by N₀ . Right ?

Okay I divided both sides by No and then took the natural log of both sides. Afterwards I divided by -5730 to get my answer, correct?

 

[SammyS]

Okay I divided both sides by No and then took the natural log of both sides. Afterwards I divided by -5730 to get my answer, correct?

Yes, but 5730 is in units of years. That gives you the decay constant in units of years^-1. MKS would be in units of seconds^-1.