# Homework Help: Radioactive Initial Decay Rate

1. May 20, 2013

### MarcL

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

The nuclide (32,15)P undergoes Beta - decay with a half-life of 14.3 d. It is used as a tracer isotope in biochemical analysis. What is the initial decay rate of a 1.2mg sample?

2. Relevant equations

T1/2 = Ln 2 / λ
Ro = λN0

3. The attempt at a solution

I first solved for λ using Ln 2 / T1/2 which gave me .04847/days.
I then converted my answer to seconds which gave me 5.61017*10^-7 /sec
I then Converted my initial amount ( 1.2mg) using U to Kg which in turn gave me 1.9926*10^21 ( using the conversion 1.6605*10^27u = 1kg)
Finally I multiplied both answer together which gave me 1.117882474*10^15 decay /sec ( the unit is given to me).
I get the answer wrong and I double checked my answer ( even triple because another question was the same thing with different number and I still got it wrong) . I don't know where I am going wrong.I would love to understand but apparently I can't even plug in formulas -.-

2. May 20, 2013

### Staff: Mentor

Your conversion factor u<->kg is wrong.
In addition, did you consider that you do not have hydrogen? The mass of an atom is not 1 u here.

3. May 20, 2013

### MarcL

Do you mean wrong in the constant I used or just the number that came out?
And are you telling me I have to find the mass of hydrogen + the mass of the neutron to then have the total mass of the nucleus?
I'm sorry, I'm kinda new at this stuff.

4. May 20, 2013

### Staff: Mentor

That is wrong. Check where the numeric factor and exponents have to be.

You can assume that [the number of protons plus the number of neutrons] in your phosphorus nucleus is its mass in u, that is a good approximation.

5. May 20, 2013

### MarcL

Well sorry I forgot to mention, when I converted my mg (well kg) to u, I then multiplied by 32 ( the atomic mass right?). For the conversion I just checked online it also gives me 1u = 1.6605*10^-27 kg ( well with more decimals).

6. May 20, 2013

### Staff: Mentor

Why did you multiply? Do you expect more atoms per kg if you have heavier atoms?

This is a different formula.

It is easier to see this with smaller numbers:
If 1A = 2*101 B, then A=20B, but not 2*10-1A = B, which would be 0.2 A = B or A=5B.

7. May 20, 2013

### barryj

1. 0.04847 is not correct.
2. why don't you express the initial decay rate as mg/sec or maybe g/sec.
Mentor, why do we care about the number of protons or neutrons??

8. May 20, 2013

### MarcL

Barry, I just input Ln 2/ 14.3 days which gave me .04847 /day no? Unless I got the formula wrong. Plus I wish I could just do mg/sec but I've been asked to do decay/sec and honestly I haven't covered that in class so I don't know what it stands for. I'm just trying by seeing other example similar to mine :/

9. May 20, 2013

### barryj

I thought they asked for the decay rate?

does the decay have to be in protons and neutrons or will grams/sec be OK.
Of course, you could convert grams into atomic particles if you like but is this what is wanted? Just asking.

10. May 20, 2013

### barryj

Ah, since it is Beta decay, there is no mass change so you would have to convert into something like Betas/sec.

11. May 20, 2013

### barryj

Beta decay is where a neutron is converted into an electron and a proton. The electron is emitted, the mass of the atom stays the same, but the atomic number is increased by 1. Is this what you have been taught?

12. May 20, 2013

### MarcL

SOrry I didn't answer. I walked away from the computer needed a break haha. Well It asks me for the rate in decay/sec and I have no idea what the decay stands for. I have been thought beta decay I just didn't thought it changed much to the mass. I thought only alpha decay changed the mass of an atom. And I just noticed I'm suppose to add a proton! I think that's where I might have been wrong.
However for my .04847/day it was only for the half life to find the disintegration constant :)

13. May 20, 2013

### MarcL

Oh I'm guessing decay per second is in bq?

14. May 20, 2013

### barryj

Check your equation for how you calculated 0.04847. Did you leave something out?
Do you remember about moles and Avogadros number. This might be appropriate.

15. May 20, 2013

### MarcL

I got the answer following an example in my book... However I don't understand the conversion it did.

It first took the atomic mass (32u in this case right?) than multiplied by mass of a single atom. In the book they used 1.661*10^-24 g/u so I used that. However I have NO idea where that comes from. Then I divided the amount of my sample ( .0012g) by the answer to the conversion above which = 2.24955*10^19.
Finally, I used the same disintegration constant I found earlier.

Can anyone understand how was that method right? :/

16. May 20, 2013

### barryj

You are getting there. What you have calculated is the initial number of atoms of Phosphorous. So, if you start with this number, and apply the delay factor, you should get the correct answer.

1.661E-24 is the mass of a proton or a neutron (approximately). so if you multiply this number by the sum of protons and neutrons, 32, you get the mass of a Phosphorous atom. Since you start with 0.0012 grams of Phosphorous, you can determine the number of phosphorous atoms in your initial; sample.

17. May 20, 2013

### barryj

I should point out that the mass does not decrease but the number of atoms that are available to decay from P to S decreases exponentially.

18. May 20, 2013

### MarcL

Ah.... I see then I get my amount of atoms initially. Does the beta decay change anything to this problem? Or is it just to confuse the reader?

19. May 20, 2013

### Staff: Mentor

Decay rate is the number of atoms to decay per second. The type of decay does not matter here.
32u is the mass of a single atom.
It is a measured value. In theory: take 10^24 hydrogen atoms, put them on a scale, and it will show 1.661g. That is not as "easy" as it sounds, and the most precise measurements use other methods, but it shows the basic concept.
You can convert this to 1.661*10^-27 kg/u, or use the inverse value of 6.02*10^26 u/kg.

[strike]I get 2.25*10^20 instead of 2.25*10^19: WolframAlpha query ("12milligram/(32u)")[/strike]
Edit: Sorry, my mistake, see the following post.

My nick is mfb :(.
It is relevant to find the number of atoms in the sample.

Last edited: May 20, 2013
20. May 20, 2013

### barryj

I think it should be 1.2 milligrams not 12 mgrams. I think 2.25E19 is correct.

I think bringing the Hydrogen atom into the discussion is just confusing the issue.