### Author Topic: physics homework  (Read 1198 times)

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#### Bobboau

##### physics homework
ok, I've tried this problem a few times but I'm not comeing up with the answer posted as the right answer.

two iron balls (mass 5.6g) are 5cm apart. if one in every billion electrons was removed from each what would be the magnatude of the electrostatic force?

my line of reasoning is thus:

with there being 26 protons (and electrons) and30 nutrons in an iron atomand the masses for protons, nutrons, and electrons being 1.67e-27(Mp), 1.68e-27(Mn), and 9.11e-31(Me) kg respectively (e is a shorthand for *10^ used by my calculator), then the mass of an iron atom must be 26*(Mp+Me)+30Mn wich I figure to be 9.384e-26kg.

in 5.6g, or .0056kg, there should be .0056/9.384e-26 = 5.96737e22 atoms of iron.

one in a billion of that is 5.96737e13, wich is the number of removed electrons. the charge of these removed electrons (in columbs (6.242e18 protons)) is -9.56e-6C, and thus the charge of each ball is 9.56e-6C.

useing coulomb's law the electrostatic force can be obtained;
Fe = k|Q1*Q2|/r^2
Fe=k|9.56e-6^2|/5^2
Fe=3.28e-2N.

wich is slightly less than 225000N.

I think I made a stupid conversion mistake somewere. someone want to help me out?
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#### karajorma

##### Re: physics homework
Hmmm. It's a physics problem so I guess working out the number of electrons removed using Avagadro's Number and the atomic mass of iron would be considered cheating right?

I can easily use that to check the first part of your calculation though.

number of atoms = number of moles x Avogadro's Number.

Number of moles = mass / RMM = 5.6/56 = 0.1 mol

Given that Avogadro's number is 6.022x10^23 you can see you're already out after the first step.

Damn. It's been ages since I did anything chemistry related so I hope I got that all correct « Last Edit: August 23, 2006, 04:03:12 pm by karajorma »
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#### Wanderer

##### Re: physics homework
In chemistry... it would be technically wrong to assume all of the iron to be of the exactly same isotope and to calculate the exact mass like in the first post.... And also the mass of electrons is neglible compared to any atomic nuclei and is generally never used for anything..

Also did you notice that the distance is in centimeters not in meters.
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#### Bobboau

##### Re: physics homework
I was makeing that assumtion, this isn't chemistry, it's physics, and the directions seemed to sugest I make that assumtion (it's actualy steel balls, but it says that steel is basicly iron and iron has 26 protons and 30 nurtons, so it seemed to me like that was the assumtion they wanted me to take).

and I did notice just after posting that it was useing centimeters when the formula wants meters, wich does give me an answere in the hundreds, rather than hundreths, but I'm still sevral orders of mangnitude off

(assumeing a pure sample of a single isotope) if you take the masses of the protons,nutrons and electrons multiplyed by how many there are your should get the mass of a single atom, and if you divide the mass of the object by the mass of the atom it should give you the number of atoms, right?
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#### karajorma

##### Re: physics homework
Bob. I said that using Avogadro's number would probably be cheating in a physics question but at least you could use it to identify whether the error lay in the first or second half of your calculation. In chemistry... it would be technically wrong to assume all of the iron to be of the exactly same isotope and to calculate the exact mass like in the first post.

I doubt I'd get the answer very wrong using 56 instead of 55.85. It was certainly all I needed to use for an order of magnitude calculation.
« Last Edit: August 23, 2006, 04:27:02 pm by karajorma »
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#### Kazan

##### Re: physics homework
remembe that both masses have the same charge after this if they started out with the same charge so make sure your force is the right sign PCS2 2.0.3 | POF CS2 wiki page | Important PCS2 Threads | PCS2 Mantis

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#### Herra Tohtori

##### Re: physics homework
Well... the number of electrons in the balls depends on what isotope they consist of.

If they are made solely of the most common iron isotope, Fe-56, then yes, there are

26 protons à 1.0072765 u,
30 neutrons à 1.0086650 u and
26 electrons à 5.4857990x10^-4 u

per one atom.

This kind of an atom, however, does not simply weigh just as much as the components put together. Together, the components weigh less than the sum of their masses, because some energy is stored into forces that keep the nucleus together. So you can't just sum the neutrons and protons together, because a nucleus formed by them weigh less than that. You have to either know the average binding energy of individual nucleods, or then you can check the literature value for Fe-56 isotope's atomic mass, which is...

M (Fe-56) = 55,85 u

When if you simply sum the particles together, you would end up presuming that the mas sper atom would be ~56.463 u.

I think there's the slight mistake you made. Just calculate again, with slightly less mass per atom, and you end up with more atoms and more electrons removed, and thus you increase the result for force, and it should go right.

EDIT: Actually the bigger error is this:

Quote
in 5.6g, or .0056kg, there should be .0056/9.384e-26 = 5.96737e22 atoms of iron.

one in a billion of that is 5.96737e13, wich is the number of removed electrons.

It's not, because there is 26 electrons per atom.

Mass per atom (presuming it's all Fe-56) is 55.85 u = 9.274117017x10^-26 kg.

Mass of a single ball is 5.6x10^-3 kg.

Atoms per single ball are

5.6x10^-3 / 9.274117017x10^-26 = 6.0383106981x10^22

Amount of electrons is 26 times that amount, 'cause there's 26 electrons per atom, thus...

1.56996078153x10^24 electrons in total/ball.

Of those, one billionth are removed. That's

1.56996078153x10^24 / 10^9 = 1.56996078153x10^15 electrons per ball removed.

So, each ball is charged at 1.56996078153x10^15 e.

e = 1.6021773x10^-19 Coulombs, so the charge per ball is

[EDIT2: Fixored line] Q = 2.51535552606x10^-4 C

From which you can easily calculate the force. Which is outwards from each ball. Any good? I'm actually getting 227457.39794 N as a result. but that's in the right magnitude all right, so it's probably just a matter of how it's calculated, how accurately things like mass loss in nuclei and the isotope configuration are taken into account.

« Last Edit: August 23, 2006, 05:04:43 pm by Herra Tohtori »
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#### Bobboau

##### Re: physics homework
oh, crap! that's it (I think) I was thinking one electron per every billionth atom there, when I should have been thinking total number of electrons!

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#### Shade

##### Re: physics homework
Yeah, if it simply refers to iron as iron and nothing more, just be creative and pick a different isotope to make the numbers fit Report FS_Open bugs with Mantis  |  Find the latest FS_Open builds Here  |  Interested in FRED? Check out the Wiki's FRED Portal | Diaspora: Website / Forums
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#### Bobboau

##### Re: physics homework
that was definately it, useing my method I get 222365.blahblahblah N. wich is the answer I'm going with because my teacher likes it when her students come up with slightly diferent answers than the highly aproxamated textbook answers. and I'm within 1/1000th of the stated answer, wich is were she usualy says I'm insane for going that precise.
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#### Eishtmo

##### Re: physics homework
Glad you figured it Bob.  That should be close enough, I would think.  Make sure you show all your work and you shouldn't have any problems.

And guys, the number of electrons are not effected by the isotope of the atom.  Isotope means different numbers of neutrons in the atom, it doesn't effect the electrons.
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#### Herra Tohtori

##### Re: physics homework
And guys, the number of electrons are not effected by the isotope of the atom.  Isotope means different numbers of neutrons in the atom, it doesn't effect the electrons.

That is very clear...

However, the isotope changes the amount of electrones in a certain amount of mass of matter. This is why:

One normal hydrogen atom consists of a proton and an electron. It has a mass of 1.008 u (atomic mass unit). That's close to one u, and it's sufficient accuracy for this purpose. So, naturally, if we have some amount of hydrogen that has mass of 1000 u, we can say that there is about 1000 hydrogen atoms in that mass, so there's about 1000 protons and 1000 electrons.

Now, an isotope of hydrogen, deuterium, consists of a proton, a neutron and an electron. It has atomic weight almost twice as big as normal hydrogen, and it is sufficiently accurate for demonstration purposes if we simply assume it to be twice the mass of normal hydrogen.

So, now we estimate deuterium atom to have a mass of 2.016 u.

If we still want to have the same mass of hydrogen, we only need about 500 deuterium atoms. 500 * 2 u = 1000 u.

So, 1000 hydrogen atoms have a combined mass 1000 u (approx.), but 500 deuterium atoms are enough to have the same mass.

Thus, same masses of different isotopes consist of different amounts of particles. The heavier the isotope, the less there are electrons in relation to lighter isotopes. Of course individual atoms of same chemical elements contain same amount of protons and electrons (assuming we're not talking about ions or plasma...), but in this case the mass was told.

Bobboau, if you wanted, you could use the given answer as a basis to approximate the isotope configuration of those iron balls of yours.
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