Decay Constant

OP

Original post by YThursday

This is a good definition:
'decay constant: The constant ratio for the number of atoms of a radionuclide that decay in a given period of time compared with the total number of atoms of the same kind present at the beginning of that period.'

I was looking at this just earlier today and came upon this forum post:

http://www.sciforums.com/showthread.php?119299-decay-constant-definition

It pretty much asks the same question :smile:

In that case, does that mean the decay constant can't be bigger than 1? Otherwise the number of atoms decaying in a time period would exceed the number of atoms at the start. And that's impossible?

Reply 4

phenomenon.ahmad

1

It most certainly can be greater than 1... we have elements with half lifes of 0.00x orders even. Seeing decay constant as a probability is the confusion here...

Reply 5

Original post by Namige

In that case, does that mean the decay constant can't be bigger than 1? Otherwise the number of atoms decaying in a time period would exceed the number of atoms at the start. And that's impossible?

That's incorrect. The examiners are deceptively astute in defining terms. It pays to credit their intelligence when considering questions/mark schemes. I've only just learnt that, lol!

The activity over a non-negligible time period is varied. Mathematically, activity is only an instantaneous quantity. In theory, an infinitesimally small quantity of time is needed to ascertain the exact instantaneous activity of a particle at any given time should the quantity be calculated by deducing a linear mean calculation. However, radioactive decay is spontaneous and random. Hence, sufficiently large periods of time and probability distributions are used to ascertain the activity at any instant. The uncertainty diminishes as the time period of measurement tends to infinity.

Also, to resolve your original problem, physicists may consider the decay constant to be a probability due to the fact that in a mathematical context, the decay constant is the expected value of the activity for 1 un-decayed nucleus (Actually, that's actually very clever of the examiners.) So (excuse my vocabulary if incorrect), think of a probability density graph (In a mathematical statistics context). You know the one i'm one about.

(edited 10 years ago)

Reply 6

YThursday

11

Original post by Namige

In that case, does that mean the decay constant can't be bigger than 1? Otherwise the number of atoms decaying in a time period would exceed the number of atoms at the start. And that's impossible?

Did you even look at the link? The guy who asks the question in that forum asks the same thing about it being able to be >1...

Reply 7

OP

Original post by YThursday

Did you even look at the link? The guy who asks the question in that forum asks the same thing about it being able to be >1...

It's far too convoluted for me to understand as I am only doing A levels. I just want someone to explain it in the simplest possible manner.

Reply 8

Original post by Namige

It's far too convoluted for me to understand as I am only doing A levels. I just want someone to explain it in the simplest possible manner.

Basically, it's called probability due to naming differences in maths and physics. In reality it's actually the expected mean average activity for 1 undecayed nuclei (when N=1, A=lambda). Therefore, of course it can be greater than 1.

You're thinking that just because the activity per unit second is greater than the actually number of undecayed nuclei, it can't happen. but that's not true. Think of emptying a bottle... in an insanely small amount of time, the contents may be pouring out a 100000000 Meters^3 per second. However, if the activity depends on the volume of water remaining (as in radioactive decay) then it will be such that the rate of expulsion will change to a small amount once the volume is small. So the contents will never actually be fully emptied.

Reply 9

OP

Original post by hecandothatfromran

Basically, it's called probability due to naming differences in maths and physics. In reality it's actually the expected mean average activity for 1 undecayed nuclei (when N=1, A=lambda). Therefore, of course it can be greater than 1.

You're thinking that just because the activity per unit second is greater than the actually number of undecayed nuclei, it can't happen. but that's not true. Think of emptying a bottle... in an insanely small amount of time, the contents may be pouring out a 100000000 Meters^3 per second. However, if the activity depends on the volume of water remaining (as in radioactive decay) then it will be such that the rate of expulsion will change to a small amount once the volume is small. So the contents will never actually be fully emptied.

So it seems what you're saying is that the rate changes depending on how much water there is. ie. dV/dt proportional to V. Then why is the decay constant: a constant?

Reply 10

Original post by Namige

So it seems what you're saying is that the rate changes depending on how much water there is. ie. dV/dt proportional to V. Then why is the decay constant: a constant?

the rate of flow of water is analogous to the activity of decay. The decay constant would be the constant of proportionality between the volume left and the rate of flow.

maybe we're not on the same page. What is it exactly you're confused about?

Reply 11

OP

Original post by hecandothatfromran

the rate of flow of water is analogous to the activity of decay. The decay constant would be the constant of proportionality between the volume left and the rate of flow.

maybe we're not on the same page. What is it exactly you're confused about?

Just the whole concept in general. So the definition 'probability of a nucleus decaying per unit time'. If the decay constant > 1, and the unit is s^-1, then if we had a second to play with the probability is more than certain that it would decay? I'm sorry if you or anyone else has already answered this, but I'm just so incredibly confused right now... The probability should only approach 1 as t approaches infinity? Assuming it hasn't decayed yet, obviously.

I understand the analogy about the water. So I see why it CAN be bigger than 1. But the definition is what confuses me.

(edited 10 years ago)

Reply 12

Original post by Namige

Just the whole concept in general. So the definition 'probability of a nucleus decaying per unit time'. If the decay constant > 1, and the unit is s^-1, then if we had a second to play with the probability is more than certain that it would decay? I'm sorry if you or anyone else has already answered this, but I'm just so incredibly confused right now... The probability should only approach 1 as t approaches infinity? Assuming it hasn't decayed yet, obviously.

I understand the analogy about the water. So I see why it CAN be bigger than 1. But the definition is what confuses me.

Ah, I think i see where your coming from. If the decay constant was defined in a mathematics exam, you couldn't call it a probability. You, and most people, are only aware of defining probability as the likelihood of an event ( which is always less than or equal to one). This is the mathematical definition and not the physics definition. In a mathematics exam, the decay constant would be able to be defined as the mean activity for 1 radioactive nuclei (I.E, when N=1, A=lambda*1). If you've done S1 and or S2, think of it as E(activity) when considering one nucleus. Since this is derived from a probability distribution, physicists may define it as a probability (although mathematically erroneous).

If you had a second to play with, it would not by any means decay with certainty. Remember that radioactive decay is random and spontaneous. Therefore, the decay constant and the exponential decay curves are estimations that attempt to predict the decay rate. It's like throwing a coin. No matter how many finite times you through it, you're not guaranteed to obtain a 50:50 split. This can only be guaranteed if the coin is thrown an infinite number of times (the detail of this, however, is deeply intuitive though).

(edited 10 years ago)

Reply 13

OP