I&#x27;m stuck on a Nuclear Decay question

Original post by NatKH

I'm given the initial activity to be 3000Bq with a half life of 6 hours. It's asking me for the number of unstable nuclei.

Unless I'm being silly, this doesn't seem to be a complete question. Can you put up the original wording please?

Reply 2

OP

Original post by atsruser

Unless I'm being silly, this doesn't seem to be a complete question. Can you put up the original wording please?

Initial number of unstable nuclei = ?
Initial Activity = 3000Bq
Half Life = 6 Hours
Decay Constant = ?
Number of unstable nuclei left after 6 hours = ?
Activity after 6 hours = ?

What is the initial number of unstable nuclei

The ones with the question marks are the ones I have to work out.

Reply 3

Original post by NatKH

Initial number of unstable nuclei = ?
Initial Activity = 3000Bq
Half Life = 6 Hours
Decay Constant = ?
Number of unstable nuclei left after 6 hours = ?
Activity after 6 hours = ?

What is the initial number of unstable nuclei

The ones with the question marks are the ones I have to work out.

If that is the literal and complete wording of the question, then I have no idea. Surely there must be more to it than that?

Reply 4

Original post by NatKH

I'm given the initial activity to be 3000Bq with a half life of 6 hours. It's asking me for the number of unstable nuclei.

It also asked me for decay constant which I got to be 3.2x10^-5 by using ln2/(6x3600) which was correct.

I'm on IsaacPhysics so when I type the answer in and I'm told if my answer is correct or not. Can anyone help with finding the number of unstable nuclei?

The relationship A = lambda N is included in the formula booklets for Physics A Levels (where A = activity, lambda = decay constant, N = number of nuclei) - this is what you need!

Reply 5

OP

Original post by atsruser

If that is the literal and complete wording of the question, then I have no idea. Surely there must be more to it than that?

That's all I'm given. Which is probably why I'm stuck on it :frown:

Reply 6

OP

Original post by Pangol

The relationship A = lambda N is included in the formula booklets for Physics A Levels (where A = activity, lambda = decay constant, N = number of nuclei) - this is what you need!

That's what I've been using but it keeps saying it's the wrong answer. I'd do 3000/3.2x10^-5 = 9.4x10^7
But it says 9.4x10^7 is wrong.

Reply 7

Original post by Pangol

The relationship A = lambda N is included in the formula booklets for Physics A Levels (where A = activity, lambda = decay constant, N = number of nuclei) - this is what you need!

Ah right. This is a problem of terminology then (to me at least). I wouldn't typically refer to N as the "number of unstable nuclei".

Reply 8

Original post by NatKH

That's what I've been using but it keeps saying it's the wrong answer. I'd do 3000/3.2x10^-5 = 9.4x10^7
But it says 9.4x10^7 is wrong.

What does it claim that the answer is? This looks OK to me.

Reply 9

Original post by atsruser

Ah right. This is a problem of terminology then (to me at least). I wouldn't typically refer to N as the "number of unstable nuclei".

Out of interest, what would you describe that N as meaning?

Reply 10

Original post by Pangol

Out of interest, what would you describe that N as meaning?

I would just refer to it as the number of atoms remaining of the radioactive species in question. Or maybe "parents", depending on the context.

Reply 11

OP

Original post by Pangol

What does it claim that the answer is? This looks OK to me.

Oh my goodness I found the answer :colondollar:

. It was 9.3x10^7 but what I did was round the decay constant before so it changed the final answer. Thank you! I couldn't look at what the answer was before because it would just tell me if I was wrong or right.

Reply 12

Original post by atsruser

I would just refer to it as the number of atoms remaining of the radioactive species in question. Or maybe "parents", depending on the context.

But if they are the atoms of the radioactive sample we are considering, then by definition, they are unstable...

Reply 13