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Silver Pin Charged

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data

    (a). Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol.
    (b). Imagine adding electrons to the pin until the negative charge has the very large value of 1.00 mC. How many electrons are added for every 10^9 electrons already present?

    3. The attempt at a solution
    I got the answers for both (a) and (b) (answers in back of book), but I still don't understand how to get (b). Heres my steps:

    For (a), all I did was dimensional analysis to find the amount of electrons in 10 g of Ag, whic is about 2.62 x 10^24 e-. For part (b), I am confused, but what I found out how many electrons in 1 mC, which was 6.25 x 10^24 e-, and divided it by 2.62 x 10^24 e-, and I got 2.38 e- for every 10^9 electrons already present, but still this does not make sense to me. Can someone explain it to me or give me some helpful hints?
     
  2. jcsd
  3. Feb 6, 2009 #2
    How many elementary electrical charges are there in one milliCoulomb?
     
  4. Feb 6, 2009 #3
    6.25 x 10^24 e-... so what do I do with that?
     
  5. Feb 6, 2009 #4
    How many groups of 10^9 electrons are there in the silver pin?
     
  6. Feb 6, 2009 #5
    2.62 x 10^24 e-/10^9 e-=2.62 x 10^15
     
  7. Feb 6, 2009 #6
    So if I divide the number of electrical charges in one milliCoulomb by the number of groups of 10^9 electrons, won't that supply the answer to the question?
     
  8. Feb 6, 2009 #7
  9. Feb 6, 2009 #8
    It seems to me that's exactly what they are asking for...

     
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