# Homework Help: Dissociation field ?

1. Feb 25, 2005

### quasar987

dissociation field ?!?

The average distance between the atomic components of a simple molecule is of the order of a few Angström. Estimate the magnitude of the field necessary to dissociate a molecule of $N_2$.

With no other information, is it possible to solve this ?!? We have a little document that talks about the atmosphere as an electrical circuit and it explains that lightning happens because the electric field dissociate molecules and then, blahblahblah, but that is IT.

2. Feb 25, 2005

### dextercioby

1.What is the dissociation energy...?Numerical value...
2.HINT:$U=\frac{qE}{d}$

Daniel.

3. Feb 25, 2005

### quasar987

1. You tell me !

I did a quick Google search.. didn't find it.

4. Feb 26, 2005

### dextercioby

Then approximate it to 1eV (order of magnitude) and then solve for E...

Daniel.

5. Feb 26, 2005

### quasar987

Cool.

.............

6. Feb 26, 2005

### quasar987

2. May I ask how you derived this formula?

7. Feb 26, 2005

### dextercioby

Sorry,it was typo in that formula
$$U=qEd$$

should have been the correct one.It was midnight.I was tired.

Daniel.

8. Feb 27, 2005

### quasar987

Ok, now the formula says that to free the atoms, we must do work on one of them in the exact amount as the binding energy. This makes a lot of sense. But there's one thing I don't get: In order to get the job done, we can chose any combination of E and d we like... i.e. we may chose to exert a small force but that acts on it a long time (over a long distance d) or the opposite (strong force, short distance). So we're not more advanced it seems :grumpy:

And oh yeah... if the field is uniform, the same force is exerted on both atoms of N. How are they ever going to dissociate?

9. Feb 27, 2005

### dextercioby

The energy is given to the 6 electrons from the chemical bond.That "d" is roughly taken as 1°A,as we don't really know on what distance should the field act...Anyway,i hope u realize this is just a formal & highly unrigurous calculation.You're interested in the order of magnitude of the electric field & that's it...

Do the calculation and then we can comment upon it.

Daniel.

10. Feb 27, 2005

### Gokul43201

Staff Emeritus
Does the question ask you to find the field it takes to ionize the molecule or the field to dissociate the molecule ? I think the latter is much harder to do than the former. You will most certainly ionize the molecule much before you dissociate it - in fact, I'm not sure that it would even be sensible to talk of dissociation by an electric field. Dissociation may result from multiple ionizations, but not of itself.

Last edited: Feb 27, 2005
11. Feb 27, 2005

### marlon

Indeed Gokul,
besides i really wonder what is happening here. There are way too much loose approximations here. I really wonder what this should lead to ???

Though this method of U =qEd seems correct to me, i have some questions on the approximations made here

marlon

12. Feb 27, 2005

### Gokul43201

Staff Emeritus
I might do a slightly different calculation (for the ionization - or breakdown - field). A not terrible approximation to use might be the "effective nuclear charge" $q_{eff}$ for a valence electron. For nitrogen, this is about 3.8 and the atomic radius is about 60 pm, from which you can get the potential energy of a valence electron, and from that, the ionization field. (Really, all you are estimating is the effective field at the radius of the valence electron).

Last edited: Feb 27, 2005
13. Feb 27, 2005

### marlon

Damn you, Gokul. I was gonna post the exact same thing...

I should type faster...

marlon

14. Feb 27, 2005

### Gokul43201

Staff Emeritus
But I get a number that is several orders of magnitude higher than the measured dielectric breakdown field strength.

15. Feb 27, 2005

### quasar987

This stuff you're talking about is too complicated for a first class of EM. I think I'll stick with dexter's way :tongue2:.

I see, I see. And, um, so q = -6e ?

It's written 'dissociate': "Estimate the strenght of the electrical field needed to dissociate a molecule of N2".

If, in order to dissociate the molecule, all it takes is to "eject" 6 of its 14 electrons, how does dissociation takes more energy then ionisation, where all 14 electrons must go ?!

Well we're talking about the dissociation process that happens in the atmosphere during a thunderstorm and that causes the N2 molecules to dissociate.

16. Feb 27, 2005

### Gokul43201

Staff Emeritus
You might dissociate N2, but it will only be thermally; not from an electric field.

To, ionize a nitrogen molecule, you only need remove ONE electron.

As far as I'm aware, ionization is the dominant process here, not dissociation - which may also happen, due to the heat generated.

Look up : www.iop.org/EJ/article/0034-4885/65/6/202/r20602.pdf[/URL]

From : [url]http://www.cltskywarn.org/lightning.htm[/url]
[quote]How a Lightning Bolt Forms

It is difficult to study lightning within the cloud; it is not a very comfortable environment for the scientist and his delicate instruments. But lightning to the ground can be seen and photographed with high-speed cameras, and from this scientists have learned much about the progressive build-up of a lightning flash. Here is the picture that emerges.

From laboratory studies of electrical breakdown of air, it is known that a lightning flash begins when the electric field reaches a strength of about three million volts per meter (75,000 volts per inch). What happens is that the few electrons that are always being freed by cosmic rays are pushed hard enough at this voltage that they knock other electrons out of the neutral molecules that they strike. These, in turn, are accelerated, collide with new molecules, and ionize them. Thus a veritable avalanche of electrons builds up, moving away from the negative charge in the cloud and leaving a trail of positive ions behind. This weakens the resistance of the air and pierces a path the developing stroke of lightning through the insulating blanket. [/quote]

Last edited by a moderator: Apr 21, 2017
17. Feb 27, 2005

### quasar987

Alrighty then. The field necessary to ionise the molecule is of the order of

$$E = \frac{U}{qd} = \frac{1 \ [eV]}{10^{-10} \ [m] \ e \ [C]} = 10^{10} \ [V/m]$$

This is too large. I remember reading that potentiel difference between the cloud and the earth before a thunderstorm is of only 10^6 V.

18. Mar 2, 2005

### quasar987

Anyone got an explanation for this?