# Homework Help: Beta+ Decay

1. Apr 23, 2010

### Lissajoux

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

${}^{12}_{7}N$ undergoes $\beta^{+}$ decay.

i. Write the corresponding equation.

ii. Calculate the decay energy, Q.

2. Relevant equations

$M\left({}^{12}N\right)=12.018613u$

$m_{e}=0.511\frac{MeV}{c^{2}}$

3. The attempt at a solution

i. For the equation:

$${}^{12}_{7}N\larrow{}^{12}_{6}N^{-}+{}^{0}_{+1}e^{+}+v_{e}$$

ii. Calculation of decay energy:

$$Q=M(12,7)-{M(12,6)+2m(e)}$$

Then input given values into this formula:

$$Q=(12.018613 \times 931.494MeV)-{M(12,6)+2(0.511MeV)}$$

.. how do I distinguish between M(12,7) and M(12,6)? Need to find
the value of M(12,6) which I think gives all values in the equation
for the calculation of Q.

I'm really not sure :grumpy:. Hopefully this is somewhat along the correct method.

2. Apr 23, 2010

### nickjer

Don't you need electron capture for this decay to occur? Also, since a proton is changing to a neutron, you will get a whole new element.

3. Apr 23, 2010

### Lissajoux

Just noticed something messed up with the equation in my initial post. Yes it is $\beta^{+}$ decay, or electron capture.

$${}^{12}_{7}N+{}^{0}_{-1}e^{-} \implies {}^{12}_{6}N^{-}+v_{e}$$

.. is this correct?

4. Apr 23, 2010

### nickjer

So the proton decays into a neutron and positron. But the isotopes will stay neutral.

$${}^{12}_7\text{N} \rightarrow {}^{12}_6\text{C}+{}^{0}_{1}e^+ + \nu_e$$

Also, remember that the element will change since the proton number changes.

EDIT: I gave the positron a "mass number" of 0 and "atomic number" of +1 to make the math easier.

5. Apr 23, 2010

### Lissajoux

Right so your equation there is correct in this case? As opposed to my equation.

How to I go about calculating the decay energy?

6. Apr 23, 2010

### nickjer

The decay energy is just the difference in mass between the parent and the daughter atoms and particles.

7. Apr 23, 2010

### Lissajoux

.. so the difference between ${}^{12}_7\text{N}$ and the sum of ${}^{12}_6\text{C}+{}^{0}_{1}e^{+} + \nu_e$?

So I'd need to look these values up online?

8. Apr 23, 2010

### nickjer

Ok, so let me rewrite this so it makes more sense. Lets pretend there are no electrons present and we just have nuclei for the isotopes and the positron. So let me write it like:

$${}^{12}_7\text{N} \rightarrow {}^{12}_6\text{C}+e^+ + \nu_e$$

Where those are individual nuclei with +7 and +6 charge, respectively (charge is now conserved). So the decay energy will be:

$$\Delta m = (\bar{m}_C+m_e) - \bar{m}_N$$

Where $$\bar{m}_C$$ and $$\bar{m}_N$$ are the masses of only the nuclei with no electrons. So lets add and subtract 7 electrons into this equation:

$$\Delta m = (\bar{m}_C+m_e) - \bar{m}_N + (7m_e - 7m_e)$$

Rearranging the masses of the electrons to form atomic masses we get:

$$\Delta m = (\bar{m}_C+6m_e) - (\bar{m}_N+7m_e) + 2m_e$$

Now writing it as atomic masses (not to be confused with the nuclear masses) we get:

$$\Delta m = m_C - m_N + 2m_e$$

So you just need to know the masses of the individual atoms, and the mass of an electron.

9. Apr 24, 2010

### Lissajoux

OK.

So I only know this though $M\left({}^{12}N\right)=12.018613u$

And the mass of an electron $m_{e}=0.511 MeV/c^{2} [/tex] As both of these are defined in the question. How do I find [itex]m_{C}$ and $m_{N}$??

10. Apr 24, 2010

### nickjer

Well you know $$m_N$$ already, it is just the mass of the nitrogen atom (you are given this value in the beginning). Also carbon-12 has an important mass. The definition of the atomic mass unit is:

"A single atom of carbon-12 has a mass of 12 u exactly, by definition."

Hopefully you remember this.

11. Apr 24, 2010

### Lissajoux

Ah yes I do remember now.

So: $$M\left({}^{12}N\right)=12.018613u , M\left({}^{12}C\right)=12.000000u , m_{e}=0.511 MeV/c^{2}$$

Then can plug these values into the equation to get the decay energy:

$$Q=\Delta m = m_C - m_N + 2m_e = -0.018613u-2m_{e} = -18.35989MeV$$

.. but this is negative. Can I just say it's positive? Or should it be $m_N-m_C+2m_{e}$??

12. Apr 24, 2010

### nickjer

Why did you subtract the $$2m_e$$ in the last line? Also if it is negative, that just means energy is released from the system and this reaction is possible.

If it were positive, then that would mean energy would be needed initially to start this reaction. And it would be impossible to decay on its own.

13. Apr 24, 2010

### karkas

Isn't it the other way around? Plus my books also say that energy from a reaction is calculated this way: $Q=[m_{initial} - m_{final}]c^2$
doesn't this hold here?

14. Apr 24, 2010

### nickjer

Well different books might write it in different ways. In the first post, Lissajoux wrote it as final - initial, so I kept it that way.