Find the moment in time when the last atom of a radioactive substance (Subsection 1b) will decay

Find the moment in time when the last atom of a radioactive...

Key Concepts

Exponential Decay

Exponential decay models describe processes where the rate of decrease of a quantity is proportional to the current amount present. In the context of radioactive decay, this means that in each time interval, a fixed fraction of atoms decays, leading to a smooth and continuous decrease over time.

Asymptotic Behavior

Asymptotic behavior in a decay process indicates how the quantity approaches a limit—in this case, zero—as time goes to infinity. Even though the substance continuously decreases, the mathematical model shows that it never completely reaches zero in any finite amount of time, but rather gets arbitrarily close.

Discrete vs. Continuous Modeling

Radioactive decay is inherently a discrete process because it involves individual atoms, yet it is often modeled using continuous exponential functions for simplicity. This continuous modeling leads to the result that the decay process takes an infinite amount of time to reach absolute zero, even though in a practical sense, all atoms may eventually decay.

Transcript

00:01 Question 83, the goal of this question is to show this relationship with a mean lifetime of a nucleus is given by 1 over lambda, one over the decay constant.

00:13 So this is what our end goal is.

00:15 So i'm going to start with defining what it means by mean time, mean lifetime of a particle.

00:22 So here i have a definition of the sum of the lifetimes of each particle, summing from 1 to total of n number of particles.

00:31 And that divided by the total sample and not my initial state of particles.

00:38 So i can rewrite my numerator as an integral now, the summation over a range, which is equivalent to an integral.

00:47 I'm just going to call that t for the time, dn, integrating from zero to infinity.

00:56 And since this is a decay process, there's a loss with every iteration.

01:02 So i'm going to, this whole integral would be negative because again, it's a decreasing rate.

01:06 So this is negative sign, negative sign comes from in front.

01:13 Yeah, i'm integrating from zero to infinity.

01:16 I'm basically extending the hint that i gave in the question where you integrate from zero to t, and then from t to t plus dt.

01:25 So continuing that logic, you can take the integral from zero to infinity.

01:28 So some of my bounds come from.

01:31 So now i have, i'm in this integral.

01:34 One thing i can note, i have time, so maybe it would be useful to display my, integral interest of d t and since i know the relationship for activity to be d n over d t is equivalent to negative lambda times the number of particles i can rearrange what i have here in the right hand side to replace my d n with a term for d t i can substitute that into my equation so i have a negative sign so my negative sign initially goes away again integrating from 0 to infinity.

02:14 My t remains, i have lambda and d t.

02:28 Additionally, i have an n over n not.

02:30 And if you look at our initial, or not initial, but our dk equation, i can rearrange this.

02:40 So i have n over and not like i have in my function there to be equivalent to negative e to the negative lambda t.

02:47 So i'll replace that ratio here.

02:53 So again, 0 to infinity.

02:57 Lambda is constant terms, i can pull that out.

03:00 So i have t, 10xpon, e negative lambda t, d t.

03:20 So i have this expression now, then i want to integrate from zero to infinity.

03:24 I think i can do this, but i, however, this is a well -known solution.

03:28 So i can just look -up, i can just access a look -up table to determine what this function would be.

03:35 So my lambda return remains out in front...

Please give Ace some feedback