# Half and mean life of nuclei

1. Aug 18, 2014

### Apollo14LMP

1. The problem statement, all variables and given/known data

Radioactive nuclei are produced in an irradiated sample at the rate of 10 s-1. If the number in the sample builds up to a maximum of 1000, calculate the mean life and the half-life of the radioactive nuclei

Can anyone advise on this one not sure how mean life is found or if half life is right .. thanks

2. Relevant equations

Using dN / dt = P - λ N

3. The attempt at a solution

1000 nuclei / 10s-1 = 100 nuclei s-1

P - λ N = 100 nuclei s-1

(In2) / λ = (In2) / (10 s-1) = 6.93 x10^-2 s-1

is Half life = 6.93 x10^-2 s-1 ??

2. Aug 18, 2014

### Orodruin

Staff Emeritus
You were given the production rate P and the N for the steadystate. What is true for the steady state?

3. Aug 18, 2014

### Apollo14LMP

Is the steady state is the difference between the rate of production 10s-1 and the rate of decay ?

4. Aug 18, 2014

### Orodruin

Staff Emeritus
The steady state is when the number of nuclei no longer changes.
It is the stable solution that N will tend to at large times.

5. Aug 18, 2014

### Apollo14LMP

So the steady state is the 1000 nuclei

So I would just need to use that amount and apply the decay constant equation ?? Im confused and going round in a circle ...

6. Aug 18, 2014

### Orodruin

Staff Emeritus
So how does your differential equation look like in the steady state?
What are the different terms?

7. Aug 18, 2014

### Apollo14LMP

1000 nuclei - 10s-1 = 990 nuclei s-1

(In2) / λ = (In2) / (990s-1) = 7.00 x 10^-4 s-1

8. Aug 18, 2014

### Orodruin

Staff Emeritus
No, your units do not make sense now. What is the meaning of each of the terms? What do they represent?

Alternatively, you can try to solve the differential equation and go from there, but it really is not necessary.

9. Aug 18, 2014

### Orodruin

Staff Emeritus
So to be specific
• What is dN/dt?
• What is P?
• What is N?
(Describe them in words, do not try to put numbers at this point)

What then are their values in the steady state?

10. Aug 18, 2014

### Apollo14LMP

There is a total of 1000 nuclei in the solution, the rate of increase (decrease) is 10s-1

If there are 1000 nuclei and the rate of decay is 10 per second, then the difference is 990 nuclei

Divided by by the decay constant 0.693 ..

11. Aug 18, 2014

### Apollo14LMP

dN is the total number of nuclei 1000
dt is the time the rate of production = 10s-1

Production rate 10 s-1
N is the total number of nuclei after I have deducted 10 nuclei from the total amount 990 nuclei

12. Aug 18, 2014

### Orodruin

Staff Emeritus
No, answer the questions in my previous post each in turn without attaching numbers.

You cannot subtract a rate from a number - the result will simply be nonsense.

13. Aug 18, 2014

### Apollo14LMP

dn / dt is the total number divided by time ...

P is the production rate

N is the total of dn/dt - production rate

14. Aug 18, 2014

### Orodruin

Staff Emeritus
Ok, we have identified some misunderstandings.

The differential equation describes how the total number of nuclei will vary with time. The left-hand side is the rate of change of the number N. dN/dt is therefore the rate of change in N.

Now, what may influence this rate? Well, we have production - we are producing a certain number of nuclei per time unit P. But the total number will also change due to decays. In a small time, the number of nuclei that decay will be proportional to the number of available nuclei, that is N. We call the constant of proportionality λ and thus the nuclei will decay with a rate λN. The total rate of change in the number of particles, i.e., dN/dt, is the rate at which nuclei are produced minus the rate at which they decay, thus
$$\frac{dN}{dt}= P - \lambda N.$$

Given the above, what would you say that dN/dt, P, and N are in the steady state solution? (Noting that it is the solution where the number of nuclei does not change with time.)

15. Aug 18, 2014

### Apollo14LMP

dn/dt = 1000 - 10 = 990
p = 10
n = 1000

mind block is here ...

16. Aug 18, 2014

### Orodruin

Staff Emeritus
No, again you are trying to subtract something with units s^-1 from a number.

The steady state is when the number of nuclei does not change with time. What is the rate of change for something that does not change with time?

Always write out the units. If you do not, your equations will not make sense.

17. Aug 18, 2014

### Apollo14LMP

dN / dt = 1000n /10/ 1 = 100
P - λ N = 10 – λ 1000N
(In2) / λ = (In2) / 1000) = 6.93x10^-4 seconds

Last edited: Aug 18, 2014
18. Aug 18, 2014

### Orodruin

Staff Emeritus
Let me repeat the question:

I am sorry, I do not know how to give you a stronger hint without giving you the answer.

It seems you are randomly throwing numbers into your equations without really understanding what they represent.

What condition between the production rate and decay rate must be fulfilled if the total number is not changing?

Or would you prefer trying to solve the differential equation?

19. Aug 18, 2014

### Apollo14LMP

Please dont be sorry - I have been struggling with this for a while. I am going to sleep on this, I am sure Im not too far off the mark. I am so grateful to you for steering me along .. thank you very much.

I must work through these issues and find my own way and solutions. Thank you once again for your sharing your expertise .....

20. Aug 28, 2014

### Apollo14LMP

dN / dt = 10 /1000 = 0.01t - the rate at which nuclei are produced minus the rate at which they decay

P - is the production rate = 10 s-1

N - number of nuclei = 1000

So, N(t) = N0 exp(-λt),

Where N0 = number of radioactive nuclei at t = 0.

N(t) = N0 exp (.01t)

N0 = N0 /2 and t = t1/2 So, t1/2 = (ln2) /.01t = 69.3 seconds

The half life is 69.3 seconds

Last edited: Aug 28, 2014