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Free electrons in solid sphere of copper 
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#1
Jan1713, 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 10^{10} 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 10^{10}  x = ##\frac{50μ}{e} \Rightarrow x = ##3.12 x 10^{14} electrons, with x the number of electrons removed. This does not make sense here. Many thanks. 


#2
Jan1713, 01:13 PM

Mentor
P: 11,621

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? 


#3
Jan1713, 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^{1}cm. I realise I made a stupid mistake. I now get ≈ 2.4g and 2.27 x 10^{22} atoms. Thanks for making me aware of my mistake.
Using this larger number of 2.27 x 10^{22}, 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. 


#4
Jan1713, 02:07 PM

Mentor
P: 11,621

Free electrons in solid sphere of copper
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. 


#5
Jan1713, 02:59 PM

PF Gold
P: 2,284

The charge on the electron I take to be 1.602 x 10^{19}.



#6
Jan1713, 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?



#7
Jan1713, 04:01 PM

Mentor
P: 11,621

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. 


#8
Jan1713, 04:17 PM

PF Gold
P: 2,284




#9
Jan1713, 04:26 PM

Mentor
P: 11,621

Okay. But make sure that you take N as being positive (a natural number).



#10
Jan1713, 04:33 PM

PF Gold
P: 2,284

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? 


#11
Jan1713, 04:45 PM

Mentor
P: 11,621

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). 


#12
Jan1713, 04:49 PM

PF Gold
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#13
Jan1713, 05:02 PM

Mentor
P: 11,621

If I do the calculation and don't make order of magnitude approximations for intermediate steps, I get a result that a bit less. 


#14
Jan1713, 05:07 PM

PF Gold
P: 2,284




#15
Jan1713, 05:24 PM

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#16
Jan1913, 08:32 AM

PF Gold
P: 2,284

@gneill
The amount I removed is about (2.9 x 10^{21}  3.1 x 10^{14}≈ 2.9 x 10^{21} ) electrons, right? I previously said this was 3.1 x 10^{14}, 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 ) 


#17
Jan1913, 09:33 AM

Mentor
P: 11,621




#18
Jan1913, 09:39 AM

PF Gold
P: 2,284

I see. So the key thing is that there are 2.85 x 10^{21} 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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