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Magnitude of the electric force a hydrogen nucleus exerts on electron?

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the magnitude of the electric force a hydrogen nucleus exerts on its only orbiting electron in the Bohr model?

    3. The attempt at a solution

    Fe = kQq/r^2

    Since there is one proton and one electron. Q and q are equal to each other: 1.6 x 10^-19.
    k = 9 x 10^9
    and radius = 40 x 10^-19.

    All the values are given in the question.

    The answer is 10^-7 but I got 1.7 x 10^-7.

    Why did they completely disregard the 1.7 ?
     
  2. jcsd
  3. Aug 11, 2014 #2

    Simon Bridge

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    It's called an "order of magnitude" calculation.
    Though it may be a typo - depends: who is "they"?

    Note: don't forget the units.
     
  4. Aug 14, 2014 #3
    Thank you I shall look into that! Sorry they is the people who wrote the solution manual and I will remember to include units next time!
     
  5. Aug 15, 2014 #4

    Simon Bridge

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    No worries - also remember, when you refer to the work of others, to include their names and the title of the work in question. Otherwise the reference is meaningless. ;)

    Note: if you use k=9x10^9 SI units, then your answer should be to 1 sig fig too.
     
  6. Jun 11, 2017 #5
    Isn't the radius supposed to be (5.3 x 10-11)?? How did you get (40 x 10-19)??

    I solved the problem the same way you did so (except that i switched the radius) and found that the answer came to be -0.82 x 10-7 C.
     
  7. Jun 11, 2017 #6

    Simon Bridge

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    Bohr radius (order e-11) would be a good choice, yes... you could akso have used e-10 (angstrom = order of magnitude size of an atom) OPs order e-19 meters would be inside the nucleus.

    Note. Dont forget units, and justify/critique guessed numbers using a physical reference. Youd probably get away with it is a secondary course but college usually penalises you if you use the correct value without indicating why. IRL if you do this you dont get published/paid.
     
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