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Specific charge

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data
    a cu atom (mass number, 63;atomic number, 29) loses 2 electrons formed. for the ion formed calculate its specific charge in c\kg



    2. Relevant equations
    specific charge=charge\mass



    3. The attempt at a solutionss
    specific charge = (29x1.6x10-19)/(63x1.67x10-27)

    i am obtaining 4.41x10^7 while the answer is 3.04x10^6.

    please help.

    thanks
    ben
     
  2. jcsd
  3. Dec 2, 2009 #2

    rl.bhat

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    Net charge on the atom is + 2e.
     
  4. Dec 2, 2009 #3
    thanks for your propmt reply. that means my charge will be 3.2x10^-19. however, how do i calculate my mass?

    cheers
     
  5. Dec 2, 2009 #4

    rl.bhat

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    Your mass in the solution is correct.
     
  6. Dec 2, 2009 #5
    thanks a lot.

    regards
    ben
     
  7. Mar 20, 2010 #6
    hey im new to this site so i really have no idea how this works but im trying to calculate specific charge as well.. but i still don't understand TT___TT...
    please help me :(
     
  8. Mar 21, 2010 #7

    rl.bhat

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    Hi lillost, welcome to PF.
    Specific charge means charge per unit mass, (e/
    Two electrons are removed from the copper atom, So the net charge Q on it is 2*1.6*10^-19 C.
    Mass of the copper atom is (63x1.67x10-27)kg.
    Now fine the specific charge on copper.
     
  9. Mar 21, 2010 #8
    aww thank you!! ^___^
    I'll kinda write down what im doing and then could you just tell me where i go wrong please?
    so then 2 elctrons have the charge : 2(1.60x10-19)
    Mass of 63Cu: 63(1.67x10-27)
    and specific charge is: charge/mass
    so we would have: 2(1.60x10-19)/63(1.67x10-27)
    = 3.2x10-19/ 1.0521x10-25
    = 3041535.976
    = 3.04x10*6?
    is that right?? im soo confused!! >____<
     
  10. Mar 21, 2010 #9

    rl.bhat

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    Your answer is correct.
     
  11. Mar 22, 2010 #10
    aw thank you very much!!
    i went through it with my teacher today, i get it now....
    ahh took me long enough...=__=
    thanks again :D
     
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