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Free electrons in solid sphere of copper

by CAF123
Tags: copper, electrons, free, solid, sphere
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CAF123
#1
Jan17-13, 12:39 PM
PF Gold
P: 2,284
1. The problem statement, all variables and given/known data
Use approximations to find the number of free electrons in a 4mm diameter solid sphere of copper. What fraction of its electrons have to be removed to leave a sphere with a charge of +50μC? Note that density of 29_Cu is 8.96 g/cm^3 and molar mass 63.54g/mol

Hint: Atomic copper has a single 4s electron in it's outer shell and when in metallic form, this electron becomes conducting.

3. The attempt at a solution
The distribution of copper is spherical so we can compute the volume easily. Using this and changing to appriopriate units, I get that: $$m = \frac{4\rho \pi r^3}{3} ≈ 2.4 \times 10^{-12}g.$$

Taking 1 mole to ##l_o## atoms, with ##l_o## the Avogadro Constant and using the given data about density and molar mass, in the end, I get that there exists 2.27 x 1010 atoms, which implies by the hint that each atom contains one freely conducting electron so there is also that same number of free electrons.
Can I check I am right up to here? Did I use the hint appriopriately?

Converting that free number of electrons to a charge, I get that the sphere would be at -3.64nC. I want this to be at +50μC, so setting up an eqn:

2.27 x 1010 - x = ##\frac{50μ}{e} \Rightarrow x = ##3.12 x 1014 electrons, with x the number of electrons removed.

This does not make sense here.

Many thanks.
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gneill
#2
Jan17-13, 01:13 PM
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That's a very unlikely number for the mass of a sphere of copper 4mm in diameter. And the number of atoms looks very low indeed.

Maybe you can show more of the calculations so we can check?
CAF123
#3
Jan17-13, 01:33 PM
PF Gold
P: 2,284
For the first calculation, I said $$m=\frac{4 \rho \pi r^3}{3},$$ with ρ = 8.96 g/cm^3 and r = 4mm →4 x 10-1cm. I realise I made a stupid mistake. I now get ≈ 2.4g and 2.27 x 1022 atoms. Thanks for making me aware of my mistake.

Using this larger number of 2.27 x 1022, I get that the charge is a massive 3638C (Ne, where N is the number of free electrons). This does not seem right at all?
EDIT: I ignored that 4mm was the diameter - trying this again...
EDIT2: so I get that the number of free electrons is about 2.84 x 10^21 and since this is an approximation of the o.o.m 10^21. This would give a charge now of about 455C.

In detail, using r=0.2cm now, gives V about 0.03cm^3. Multiply this by density (8.96) to give a mass of 0.3g. 1 mole to l_o atoms and 63.54g/mol implies with the given mass of 0.3g, we are dealing with about 4.73 x 10^{-3} mols. Now multiply this by Avagoadros const to get about 10^{21} free electrons. Is this correct now?

When I multiply by e, should I take e to be -ve? I.e the charge on the sphere would be about -455C, taking charge to be flow of conducting electrons.

gneill
#4
Jan17-13, 02:07 PM
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P: 11,621
Free electrons in solid sphere of copper

Quote Quote by CAF123 View Post
For the first calculation, I said $$m=\frac{4 \rho \pi r^3}{3},$$ with ρ = 8.96 g/cm^3 and r = 4mm →4 x 10-1cm. I realise I made a stupid mistake. I now get ≈ 2.4g and 2.27 x 1022 atoms. Thanks for making me aware of my mistake.

Using this larger number of 2.27 x 1022, I get that the charge is a massive 3638C (Ne, where N is the number of free electrons). This does not seem right at all?
EDIT: I ignored that 4mm was the diameter - trying this again...
EDIT2: so I get that the number of free electrons is about 2.84 x 10^21 and since this is an approximation of the o.o.m 10^21. This would give a charge now of about 455C.

In detail, using r=0.2cm now, gives V about 0.03cm^3. Multiply this by density (8.96) to give a mass of 0.3g. 1 mole to l_o atoms and 63.54g/mol implies with the given mass of 0.3g, we are dealing with about 4.73 x 10^{-3} mols. Now multiply this by Avagoadros const to get about 10^{21} free electrons. Is this correct now?

When I multiply by e, should I take e to be -ve? I.e the charge on the sphere would be about -455C, taking charge to be flow of conducting electrons.
Okay, I see you're sorting out the slips.

The free electrons (or more specifically, the conduction band electrons) of the copper atoms do not represent a net charge on the sphere; they are happily balanced by protons in the nuclei of the copper atoms, rendering each atom net neutral. It's just that these electrons can move easily from atom to atom, and is why copper is "conductive".

The question just wants to know the number of electrons in the conduction band, which is the same as the number of atoms in the sphere, which you've done.

Now, when it comes to determining the number of electrons that you need to remove in order to place a charge of 50 μC on the sphere, you'll need to know the charge on the electron.
CAF123
#5
Jan17-13, 02:59 PM
PF Gold
P: 2,284
The charge on the electron I take to be -1.602 x 10-19.
CAF123
#6
Jan17-13, 03:15 PM
PF Gold
P: 2,284
A charge of +50μC allows an electron number density to be associated. So the number of free electrons is simply ##N = q/e, \Rightarrow N = ##50 x 10-6/(-1.602 x 10-19), whereas that for about -455C is -455/(-1.602 x 10-19). (which is just the number of free electrons i calculated).Take one away from the other and divide by initial amount on sphere to get the fractional removal. Correct?
gneill
#7
Jan17-13, 04:01 PM
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Quote Quote by CAF123 View Post
A charge of +50μC allows an electron number density to be associated. So the number of free electrons is simply ##N = q/e, \Rightarrow N = ##50 x 10-6/(-1.602 x 10-19), whereas that for about -455C is -455/(-1.602 x 10-19). Take one away from the other and divide by initial amount on sphere to get the fractional removal. Correct?
The 50 μC is not directly related to the free electrons which are an intrinsic property of copper.

They're looking for the fraction that has to be removed, not the percentage change. That would correspond to the number to be removed divided by the total number to begin with.

Now, the question is slightly tricky here, because they say "What fraction of its electrons have to be removed to leave a sphere with a charge of +50μC?", after they've just asked you to calculate the number of free (conduction) electrons. From the phrasing, the "its" most likely refers to the total number of electrons that the sphere has, not just its free electrons. So you should now calculate the total number of electrons on the neutral sphere, and the number of electron charges making up 50 μC of charge.
CAF123
#8
Jan17-13, 04:17 PM
PF Gold
P: 2,284
Quote Quote by gneill View Post
So you should now calculate the total number of electrons on the neutral sphere,
That being ≈1021 multiplied by 29.

and the number of electron charges making up 50 μC of charge.
That being N = 50μC/e, taking e = -1.602 x 10-19C
gneill
#9
Jan17-13, 04:26 PM
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Okay. But make sure that you take N as being positive (a natural number).
CAF123
#10
Jan17-13, 04:33 PM
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P: 2,284
Quote Quote by gneill View Post
Okay. But make sure that you take N as being positive (a natural number).
So intially on the neutral sphere we have 2.9 x 1022 electrons. Similarly, on the 50μC sphere we have about 3.1 x 1014 electrons. This means to get it to that charge, I take about 3.1 x 1014electrons off. (2.9 x 1022 - 3.1 x 1014). Divide this by 2.9 x 1022 and I get my fractional removal?

One pedantic question: When you say make N positive, how can I do that? q is positive (+50) and e is negative. In this case, do we just say let the physics drive the maths in that the number of electrons cannot be negative?
gneill
#11
Jan17-13, 04:45 PM
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The sign of the charge on the electron isn't important; it's the magnitude. You want to know how many electron charges go into making up a charge of size 50 μC. If you remove that many electrons from the sphere, it's left with a +50 μC net charge imbalance.

The fraction of the sphere's electrons that you removed is N/(starting number).
CAF123
#12
Jan17-13, 04:49 PM
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P: 2,284
Quote Quote by gneill View Post

The fraction of the sphere's electrons that you removed is N/(starting number).
Is that not what I wrote? 3.1 x 1014/2.9 x 1022 ≈ 10-8
gneill
#13
Jan17-13, 05:02 PM
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Quote Quote by CAF123 View Post
Is that not what I wrote? 3.1 x 1014/2.9 x 1022 ≈ 10-8
Well, you described it but didn't perform the math

If I do the calculation and don't make order of magnitude approximations for intermediate steps, I get a result that a bit less.
CAF123
#14
Jan17-13, 05:07 PM
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P: 2,284
Quote Quote by gneill View Post
Well, you described it but didn't perform the math

If I do the calculation and don't make order of magnitude approximations for intermediate steps, I get a result that a bit less.
Ok, thanks for the check. Do you think it would matter since this is like an 'estimation' question? Thanks again.
gneill
#15
Jan17-13, 05:24 PM
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Quote Quote by CAF123 View Post
Ok, thanks for the check. Do you think it would matter since this is like an 'estimation' question? Thanks again.
Well, you're given some "hard" numbers to work with, and you haven't really had to estimate anything. My suggestion would be to carry out all calculations to the given number of significant figures. If they ask for an order of magnitude at the end, take it from the calculated value.
CAF123
#16
Jan19-13, 08:32 AM
PF Gold
P: 2,284
@gneill
The amount I removed is about (2.9 x 1021 - 3.1 x 1014≈ 2.9 x 1021 ) electrons, right? I previously said this was 3.1 x 1014, for some reason. (That is more or less the whole lot)

(number of electrons on orginal sphere - amount removed = number of electrons on 50μC
implies
amount removed = number on orginal sphere - number on 50μC )
gneill
#17
Jan19-13, 09:33 AM
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Quote Quote by CAF123 View Post
@gneill
The amount I removed is about (2.9 x 1021 - 3.1 x 1014≈ 2.9 x 1021 ) electrons, right? I previously said this was 3.1 x 1014, for some reason. (That is more or less the whole lot)

(number of electrons on orginal sphere - amount removed = number of electrons on 50μC
implies
amount removed = number on orginal sphere - number on 50μC )
No, that's not right. You want to remove 50 μF worth of electrons from the sphere. That's the 3.1 x 1014 number you calculated earlier. The neutral sphere starts out with about 29 x 2.85 x 1021 electrons.
CAF123
#18
Jan19-13, 09:39 AM
PF Gold
P: 2,284
I see. So the key thing is that there are 2.85 x 1021 x 29 electrons on the neutral sphere. Removing 50μC of electrons results in a net 50μC positive charge on the sphere.
Thanks.


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