How Many 116In Nuclei Can Be in Equilibrium?

[Apollo14LMP]

Homework Statement

Can someone check this please ?

Calculate the maximum number of 116In nuclei (number in equilibrium) in the sample if the half-life of 116In is 54 m.

Homework Equations

The Attempt at a Solution

The half life is 54 minutes, converted to seconds = 54m x 60 s-1 = 3.24 x 10^3 s-1

(In2) / λ = In2 / 3.24 x 10^3 s-1 = 2.14 x 10^-4 s-1

A = R - λN(t) So, A = 10 s-1 / 2.14 x 10^-4 s-1= 4.67×10^4 nuclei

The number of nuclei in the sample N = mNA/A, where is the fraction of the isotope (atomic mass A) in the sample (mass m) and NA is Avogadro's number.

N = 4.67×10^4 x 6.022^23 / 116 = 2.42 x 10^26 s-1 Equilibrium activity = 2.42 x 10^26 s-1

Homework Statement

 

[DEvens]

That sure does not look like the entire problem statement. There seems to be an implied production of 116In that you have not explained.

There are 60 seconds per minute. Your notation is weird. You seem to be getting inverse seconds instead of seconds for the seconds equivalent to 54 minutes.

Can't figure out what the various things in A = R - λN(t) mean. Can't figure out how that has anything to do with the rest of that line, since you don't subtract anything from anything. Does not look like you have calculated λ, just used an equation it appears in, so the substitution is confusing.

That's enough places where you are very sloppy. I'm stopping. Maybe you could start over and be more careful?

 

[Apollo14LMP]

In the capture reaction 115In + n 116In, 116In nuclei are produced at the rate of 10 s-1. Calculate the maximum number of 116In nuclei (number in equilibrium) in the sample if the half-life of 116In is 54 m.

The half life is 54 minutes, converted to seconds = 54m x 60 s-1 = 3240 s-1

(In2) / λ = In2 / 3240 s-1 = 2.14 x 10^-4 s-1

So, 10s-1 / 2.14 x 10^-4 s-1 = 4.67×10^4 nuclei

The number of nuclei in the sample N = mNA/A, where the fraction of the isotope (atomic mass is A) in the sample (mass m) and NA is Avogadro's number.

N = 4.67×10^4 x 6.022 x 10^23 / 116 = 2.42 x 10^26 s-1

Equilibrium activity = 2.42 x 10^26 s-1

 

[BvU]

Perhaps you didn't get DE's message; if you don't want to tell, that's fine. In the mean time you leave me wondering what you mean with N=mNA/A. Was m a given ? then why not tell us, especially after DE explicitly asks for the entire problem statement ? But I suspect m is not a given, because it's irrelevant.

How can you write N = ... s-1 ? Is s-1 the dimension of a number ?

Why care about N at all if you were given that the 116In nuclei are produced at the rate of 10 s^-1 ?

What on Earth is In2 ? Never heard of that isotope !Please spend some time reading the guidelines and the tips and what have you, so you can write s^-1 if that is what you mean with "per second" (which is NOT, repeat NOT the dimension of a half-life).

Oh, and: did you notice all your relevant equations have disappeared as if by magic ? No wonder there's nothing sensible coming out ! Please spend some time collecting and rendering the relevant equations.

This isn't meant as bashing, but really: if you want someone to help you (and there really are people about who do), you shouldn't make it impossible for them !

So start over with this simple exercise: there are ten 116In nuclei that get produced per second. If there are N 116In nuclei present, the number that decay per second is something like N/$\tau$. So by the time N/$\tau$ = 10 s^-1 just as many decay as there are being formed. What can be easier ?

 

[Orodruin]

The problem is in all aspects (apart from actual numbers and that we are here solving for a different quantity) equivalent to your earlier thread
https://www.physicsforums.com/showthread.php?t=766515
Why don't you go back and review that thread rather than starting a new one on the same subject? The replies you will get are going to be saying essentially the same as what I said there. Your main problem in that thread seemed to be interpreting what different terms were actually representing as well as unit analysis when adding different quantities.