Quantum Fluctuations inside Atoms?

[Dickie_Boy]

Hi,

In 'Nothingness' matter is apparently always popping in and out of existance (quantum fluctuations?)

Now I've always visualised this happening in the vacuum of space. Although on hindsight I don't think even the vacuum of space is totally 'Nothing'? Isn't there a stray atom or two in every cubic metre?

Anyhow, is matter popping in and out of existance inside atoms, as 90% of them is 'Nothingness'?

[JulianO]

Hi,

In 'Nothingness' matter is apparently always popping in and out of existance (quantum fluctuations?)

Now I've always visualised this happening in the vacuum of space. Although on hindsight I don't think even the vacuum of space is totally 'Nothing'? Isn't there a stray atom or two in every cubic metre?

Yes indeed - interstellar space is usually characterised as a few atoms per cubic meter. Far far better than any vacuum we can make on Earth, but not empty.

The estimate is a density of about 7×10^−29 g/cm^3, and an atom of hydrogen is about 1.66×10^−24 g - which works out at about 42 atoms per cubic meter (a suspiciously significant answer ).

Anyhow, is matter popping in and out of existance inside atoms, as 90% of them is 'Nothingness'?

Yes indeed, everywhere.

[Dickie_Boy]

Thankyou, that's amazing!

So when I say I'm a 'Nobody' I really am (apart from a little bit of matter). AND I'm permantly fizzing with quantum fluctuations :-)

[Dickie_Boy]

which works out at about 42 atoms per cubic meter (a suspiciously significant answer ).

.

That's interesting, because apparently an atom is relative to an apple, as an apple is relative to the earth (same number of atoms and apples apparently).

[JulianO]

Actually - a lot more of an atom than 90% is nothing.

The nucleus of an atom is about a 10,000th of the size of an atom. Here are some nice analogies from one of the OU's text books.

• An atom with a radius the same as the Earth’s orbit (1 AU) would have a nucleus much smaller than the Sun! Since 1 AU ≈ 1.5 × 10^11 m, the scaled nucleus would have a radius 1.5 × 10^6 m, about the same as the radius of the Moon!
  
• The continent of Australia is roughly 4000 km across. An atom this size would have a nucleus only 40 m across, about the height of Sydney Opera House.
  
• The M25 ring road around London has a diameter of order 40 km. If an atom had a radius this size, the nucleus would be only 0.4 m across, or the width of a rubbish bin in central London.
  
• An athletics stadium is typically about 100 m across, so an atom on this scale would have a nucleus about 1 mm across, or the diameter of a pinhead.
  
• As a final example, an atom the size of a house 10 m across would have a nucleus only 0.1 mm in diameter, smaller than the full stop at the end of this sentence.
  

[Dickie_Boy]

That's Doubly Amazing

As a side issue:

When 'They' split the atom do they really mean they're splitting the nucleus and breaking apart the strong and weak nuclear forces?

Also is the mass in the equation E=mc^ just the mass of the protons and neutrons and not the electrons?

I think the electrons are so squiddly small they can be ignored anyway, is that right?

[JulianO]

T

When 'They' split the atom do they really mean they're splitting the nucleus and breaking apart the strong and weak nuclear forces?

Yes - well usually the strong force, that's what holds together nucleons. The weak force is more to do with decay.

Also is the mass in the equation E=mc^ just the mass of the protons and neutrons and not the electrons?

Whatever mass you are interested in, be it a proton, a neutron or a house brick.

In a thermonuclear explosion its the difference in mass before and after. The components of a uranium atom after they are split weigh fractionally less than the original atom. For fusion you get more bang for the buck, as a He nucleus made from 4 H atoms has a lot more missing mass than the U split. It's this binding energy that you are releasing in the bomb (or power station). In this case all the mass is lost from the nucleus. The electrons would be kicked away from the atoms they were on, but would eventually find new homes.

I think the electrons are so squiddly small they can be ignored anyway, is that right?

It's often the case, as they are about 1/1800th of the mass of a proton, but it depends what you're calculating.

[Dickie_Boy]

Oh I see...

And in fssion, do they choose elements like uranium because they have a lot of mass (lots of neutrons and protons in their nucleus) and thus release a lot of energy

Or is it because they are the most unstable and easiest to split?

Or, a is it bit of both?

ALSO:

I didn't realise that the energy that was released was proprtional to the difference in the 'before and after'. Is that why new elements forms after nuclear reactions? Some of 'stuff' released clumps together forming new elements and the difference in mass in the before and after is the BANG?

[JulianO]

And in fssion, do they choose elements like uranium because they have a lot of mass (lots of neutrons and protons in their nucleus) and thus release a lot of energy

Or is it because they are the most unstable and easiest to split?

Or, a is it bit of both?

A bit of both really. Uranium is right on the edge of being stable, poke it with an extra neutron and it will tip over the edge and split. Also as far as binding energy goes, Iron is the most stable, so the further you go from there in either direction the more energy you release, either by splitting/fission (bigger nuclei) or joining/fusion (smaller nuclei).

You can see from the above how much more efficient fusion of Hydrogen is that fission of Uranium - its about 6 units from fusion of H -> He, and even if you managed to split U into Fe you'd only get about .5 units. That's why stars work best burning Hydrogen. The can burn other things, like He->C but its not nearly as good as H->He.

I didn't release the energy that was released was proprtional to the difference in the 'before and after'.

Yup - the old E=mc^2!

Is that why new elements forms after nuclear reactions? Some of 'stuff' released clumps together forming new elements and the difference in mass in the before and after is the BANG?

Well - its sort of simpler than that. Typically a U atom will split into two vaguely equal parts.

As you can see, it can happen in various ways. Often those products such as Cs-140 are radioactive too, though they don't undergo fission, just radioactive decay - although again with a small loss of mass and release of energy.

Similarly with fusion, there are about 7 or 8 different ways you can fuse 4 H to make an He. Bigger element fusing is more complex, and fusing to anything above Iron (Fe) requires you to put energy in, which is why they are typically only made in supernovae. However big stars can make some heavy elements by a slow process of addition during their normal burning - although there are only certain isotopes that can be made that way.

[Dickie_Boy]

Oh dear I think this is getting a bit beyond me...

Is it the bigger the number of MeV the more energy that is released?

Also, In the two pictures of the Uranium atom. If I add up the atomic numbers of the elements after the BANG (probably the wrong term?) In the first picture they add up to more that the atomic number of U235, and in the second picture less than the atomic number of U235? What's going on 'ere :-/

PS

Please bear with me because I will probably have some more questions when I finally get my head round what happens in nuclear fission and fusion :-?

PPS

I wish I had your brain

[JulianO]

Is it the bigger the number of MeV the more energy that is released?

Yes - MeV is a convenient measurement of small amounts of energy. (Million electron volts - you could have it in SI units - the Joule, but its a very small number then 1eV ~ 10^-19 J)

Also, In the two pictures of the Uranium atom. If I add up the atomic numbers of the elements after the BANG (probably the wrong term?) In the first picture they add up to more that the atomic number of U235, and in the second picture less than the atomic number of U235? What's going on 'ere :-/

In the fist one we have U-235 + n -> Ba-141 + Kr-92 + 3n

So thats 235+1 -> 141 + 92 +3 = 236

Second one

U-235 + 1n -> Ca-140 + Rb-92 + 2n

235+1 = 140+92+2 = 234 so that one doesn't add up I agree. I'm not even sure its a real example.

So discount that one!

The other things is that although Ba-141 has 141 nucleons in its nucleus, and you might think its atomic mass was 141 - it isn't - quite. Its actually 140.914411 - and the difference in mass is the binding energy. Its what's holding the nucleus together. You'd need to supply that difference in terms of energy equivalent to break it up into its bits again.

[Dickie_Boy]

Many thanks once again.

I've been watching a good video I found on YouTube, but I would guess you know it all already as you've mentioned most of it already. Added to what you've told me there's a lot for me to get to grips with, but I'll get there in the end, hopefully