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Atomic models and coloumbs law

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    The radius of the orbit of an electron in a hydrogen atom is 0.05 nm. Using the planetary model, calculate the orbital velocity needed by the electron to avoid being pulled into the nucleus by electrostatic attraction. Round up your answer to the nearest tenth.

    2. Relevant equations

    mv^2=k(e^2/r)
    mass of electron -13.6?? (not sure)
    v=velocity
    k= coloumb constant
    e=1.6x10^-19

    3. The attempt at a solution

    13.6(v^2)=k(q1q2/r2)(1.6x10^-19/5x10^11 m)
    i think the equation is wrong :( im not sure about coloumbs constant and what it is..
     
  2. jcsd
  3. Feb 9, 2009 #2

    Tom Mattson

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    That much is right.

    It's OK if you aren't sure of the mass of an electron, but for Jeebus' sake son use your brain. When was the last time you put something on a scale and found that its weight is negative? :tongue:

    You should look up the electron mass. That's not too much to ask.

    OK

    While you're looking up the electron mass, look up that constant as well, please. They are both in your book.
     
  4. Feb 9, 2009 #3

    LowlyPion

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  5. Feb 9, 2009 #4

    Redbelly98

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  6. Feb 9, 2009 #5
    ok. now i understand that its just asking for coloumbs constant and not for the law. but my course is online not text book, and has never showed me 8.99x10^9Nm2/C2 as coloumbs constant, all it says in the tutorial is that coloumbs constant (k) makes sure the units work out properly.
    i understand now that e= charge of electron which is 1.6x10^-19 coloumbs
    and i understand now that the mass of an electron is 9.10938188 × 10-31 kilograms but is that what its asking for in the equation? of mv^2=k e^2/r^2?
    thanks for your replies
    and i also know that r is the radius
    and v is what were looking for
    thanks for your replies
     
  7. Feb 10, 2009 #6

    Redbelly98

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    The electron mass is the m in that equation. They're not asking for m, but you need it in order to calculate what v is.

    EDIT:
    p.s. typo? The equation really is
    m v^2 / r = k e^2 / r^2​
    or
    m v^2 = k e^2 / r​
    as you had originally.

    Good luck :smile:
     
  8. Feb 10, 2009 #7
    o ok. and btw it was a typo
    m v^2 = k e^2 / r
    ok so i got everything down now im just still confused about k do i insert 8.99x10^9Nm2/C2 for k or like i said before all k does is it "makes sure the units work out properly."
     
  9. Feb 10, 2009 #8

    Redbelly98

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    Yes.
     
  10. Feb 11, 2009 #9
    how does this look?
    m v^2 = k e^2 / r
    9.10938188×10^-28(v^2)=8.99x10^9(1.6x10^-19^2/5x10^-11)
    9.10938188×10^-28(v^2)=8.99x10^9(5.12x10^-28)
    9.10938188×10^-28(v^2)=4.60288x10^-18
    v^2=5052900472
    v= 71080 m/s
     
  11. Feb 11, 2009 #10

    Redbelly98

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    The electron mass is wrong.
     
  12. Feb 11, 2009 #11
    i put the electron mass in grams i thought it was supposed to be in the basic unit.
    so is it supossed to be in kilo? 9.10938188 × 10-31?
     
  13. Feb 11, 2009 #12

    Redbelly98

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    Yes. kg, not grams, are the base mass unit in the SI metric system.

    Example: 1 Newton = 1 kg·m/s²
     
    Last edited: Feb 11, 2009
  14. Feb 11, 2009 #13
    ok. so that made a big difference in the answer
    m v^2 = k e^2 / r
    9.10938188×10^-31(v^2)=8.99x10^9(1.6x10^-19^2/5x10^-11)
    9.10938188×10^-31(v^2)=8.99x10^9(5.12x10^-28)
    9.10938188x10^-31(v^2)=4.60288x10^-18
    v^2=505290047x10^-27
    v= 7.10837567x10^-14 m/s

    how is that?
    and its meters per second right.
     
  15. Feb 11, 2009 #14

    Redbelly98

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    I will be honest. It is close to the point where I wonder if you are really trying to solve this problem, or are deliberately making mistakes so that other people (i.e. me) waste their time responding.
     
  16. Feb 11, 2009 #15
    ok thanks i appreciate your honesty.
    if you dont want to help its fine but could you at least give me a website that explains because ive been looking and i cant find anything that helps me.
     
    Last edited: Feb 11, 2009
  17. Feb 11, 2009 #16

    Redbelly98

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    If I'm wrong about my suspicions, I sincerely apologize.

    You have been so close to solving this, but made an arithmetic error somewhere in post #13. Your formula is correct. Your input values (k, m, e, r) are correct. It's just a matter of being careful with the arithmetic at this point.
     
  18. Feb 11, 2009 #17
    o wow. i can see now why you thought i was kidding, i think i got it now.
    m v^2 = k e^2 / r
    9.10938188×10^-31(v^2)=8.99x10^9(1.6x10^-19^2/5x10^-11)
    9.10938188×10^-31(v^2)=8.99x10^9(5.12x10^-28)
    9.10938188x10^-31(v^2)=4.60288x10^-18
    v^2=5.052900472x10^12
    v= 2247866 m/s
    and btw i have a ti 84 calc and for some reason whenever i enter scientific notations like 5.12x10^-28 and divide it by something or multiply, it will always give me a different answer and when i do it manually and i insert all the 0s it will give me the right answer, i dont know if its the calc, or something else that im doing wrong but its a hassle because i always have to insert so many 0s! any advice?
     
  19. Feb 11, 2009 #18

    Redbelly98

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    Yes! And I'm sorry about the misunderstanding.

    I have a ti 38+, it's probably the same as the ti 84.

    To enter 9.1x10^-31, the keystrokes are:

    9
    .
    1
    "2nd"
    "EE" (NOTE: same key as ",", just about "7" key)
    "(-)" (NOTE: not the "-" subtraction key, the "(-)" key is below the "3" key)
    3
    1

    Is that how you are doing it? By the way the "EE" stands for "Enter Exponent", it means "times 10 raised to the exponent to be entered next"
     
  20. Feb 11, 2009 #19
    o ok i think that should work i was doing it by inserting
    9
    x
    10
    ^
    -
    31
    so i think your way should work.
    thanks for the replies and for helping me out with my question
     
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