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Homework Help: Find distance electron travels before turning around

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data
    The hydrogen molecular ion H1,2, with one electron and two protons, is the simplest molecule. The equilibrium spacing between the protons is 0.11 nm. Suppose the electron is at the midpoint between the protons and moving at 1.5 * 10^6 m/s perpendicular to a line between the protons.

    How far (in nm) does the electron move before reaching a turning point? Because of their larger mass, the protons remain fixed during this interval of time.

    2. Relevant equations

    PE = kQq/r
    KE = 1/2 mv2
    W = KE + PE
    k = 9*109

    3. The attempt at a solution

    I first drew a diagram to get an idea of how to approach this problem.

    Let x2 = .055 nm

    R = √(D2 + x2)

    -ΔKE = ΔPE
    -(-KEi) = PEf

    1/2 mv2 = kQq/R + kQq/R = 2kQq/R

    R = 4kQq/(mv2) = √(D2 + x2)

    D2 + x2 = [4kQq/(mv2)]2

    D = √( [4kQq/(mv2)]2 - x2 )

    D = ~.447 nm

    This answer, however, is wrong. The correct answer is 3.0 * 10-3 nm.

    I tried integrating from F, but I still found the same equation for PE. Am I doing something incorrect? Please help as I need to know this by tomorrow for my midterm. Thank you for any help received!

    Attached Files:

  2. jcsd
  3. Mar 1, 2013 #2
    So the KE = -PE method is correct? If it is, then I'm not sure what I'm doing wrong at all.

    The final answer I have does not match up to the correct answer. It's even off by a factor of ten.
  4. Mar 1, 2013 #3


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    PE+KE = const. The potential energy is not zero at the midpoint. Change of potential energy= final KE.

  5. Mar 1, 2013 #4
    So the set-up should be

    PEf - PEi = KEf

    2KQq (1/R - 1/x) = 1/2 mv^2 ?
  6. Mar 1, 2013 #5
    What is x2 and why is it 0.055 nm?
  7. Mar 1, 2013 #6


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  8. Mar 1, 2013 #7
    x is just a value I used to clean up the work before plugging everything in.

    The problem states that the electron travels in the path of a perpendicular bisector of an imaginary line connecting the two protons. If you were to find the radius of that, then it'd be half of the length of the line squared plus the distance traveled squared under a radical. Rather than saying Sqrt( L^2/4 + D^2), it seemed cleaner to say Sqrt( x^2 + D^2) and make x = (1/2)L, L being the length of the line connecting the two protons.
  9. Mar 1, 2013 #8
    So I have something like this:

    2kQq(1/R - 1/x) = 1/2 mv2

    (x-R)/(xR) = (mv2)/(4kQ2) <--- because Q=q=1.6*10-19

    (xR)/(x-R) = (4kQ2)/(mv2)

    mv2xR = 4kQ2x - 4kQ2R

    (mv2x + 4kQ2)R = 4kQ2

    R = (4kQ2)/(mv2x+4kQ2)

    √(x2+D2) = (4kQ2)/(mv2x+4kQ2)

    D2 = (4kQ2)/(mv2x+4kQ2)2 - x2

    D = √[(4kQ2)/(mv2x+4kQ2)2 - x2]

    However, when I plug in all the values:
    x = .055 * 10-9 m
    Q = 1.6 * 10-19 C
    m = 9.1 * 10-31 kg
    v = 1.5 * 106 m/s

    I get the square root of a negative number.


    The set up is supposed to be

    KEi = PEi - PEf

    BUT it should look like this:

    -2kQq(1/R-1/x) = 1/2 mv2

    2kQq(1/x-1/R) = 1/2 mv2 because Q (the source charges) is 1.6 * 10-19 while q = -1.6 * 10-19

    (R-x)/(xR) = (mv2)/(4kQ2)

    (xR)/(R-x) = (4kQ2)/(mv2)

    mv2xR = 4kQ2R - 4kQ2x

    (mv2x - 4kQ2)R = 4kQ2

    R = (4kQ2)/(mv2x-4kQ2)

    √(x2+D2) = (4kQ2)/(mv2x-4kQ2)

    D2 = (4kQ2)/(mv2x-4kQ2)2 - x2

    D = √[(4kQ2)/(mv2x-4kQ2)2 - x2]

    Plug everything in, and you should be able to get 3.00131072 * 10^-2 nm
    Last edited: Mar 1, 2013
  10. Mar 1, 2013 #9
    But why is its SQUARE equal to 0.055 nm?
  11. Mar 1, 2013 #10
    Oh, I'm sorry. That was a typo in the introduction of the problem. It is supposed to be x = .055 nm.

    I have not been using the wrong value for x or x2 though, so I am still a little lost.


    Refer to my previous post as the answer and work is in that post.
    Last edited: Mar 1, 2013
  12. Mar 1, 2013 #11
    Is that really so?
  13. Mar 1, 2013 #12


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    It is
    2kQq(1/x - 1/R) = 1/2 mv, as KE(initial)=PE(final)-PE(initial). But qQ is a negative number, so your equation is correct if you use the absolute value for the electron charge.

    Previously you called the vertical displacement "x". It is better to denote the distance from a proton by "d", for example.

    Do not complicate the calculations. Plug in the data, calculate 1/R-1/x. Do not round too much.
    (1/R-1/x) can be also written as 1/R(1-R/x).

  14. Mar 1, 2013 #13
    Thank you. As you see above, I found the correct answer. "x" was not set as the vertical displacement, however. Instead, I set "x" as half of the length of the line connecting the two protons and "D" as the distance from the midpoint of the referred line.

    I also now see why it is (1/x - 1/R) due to your explanation. As always, you seem to have fully answered my questions without giving away the answers, allowing me to figure out the entire problem. Thank you so much for all the times you have helped me, ehild! :)
  15. Mar 1, 2013 #14
    That was my greatest error, as ehild also pointed out. Thank you for your time and effort in helping me understand this problem!
  16. Mar 1, 2013 #15


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    Well, sorry, I forgot your notations. The PE(final)-PE(initial)=2kqQ(1/R-1/x) , your equation was correct. But q is negative.
    Have you got 0.003 nm really? I could not get the same result...

  17. Mar 1, 2013 #16
    Yes I have!
    Like you said, since q is negative, the (1/R-1/x) becomes (1/x-1/R) instead.

    PEf - PEi = 1/2 mv2

    2kQq(1/R-1/x) = 1/2 mv2

    -2kQ2(1/R-1/x) = 1/2 mv2 Since q = -Q

    2kQ2(1/x-1/R) = 1/2 mv2

    1/x - 1/R = (mv2)/(4kQ2)

    (R-x)/xR = (mv2)/(4kQ2)

    xR/(R-x) = (4kQ2)/(mv2)

    (mv2) xR = (4kQ2)R - (4kQ2)x

    [(mv2)x - (4kQ2)]R = -(4kQ2)x

    R = -(4kQ2)x/[(mv2)x - (4kQ2)]

    Since R = √(x2+D2)

    √(x2+D2) = -(4kQ2)x/[(mv2)x - (4kQ2)]

    x2+D2 = [ -(4kQ2)x/[(mv2)x - (4kQ2)] ]2

    D2 = [ -(4kQ2)x/[(mv2)x - (4kQ2)] ]2 - x2

    D = √ ( [ -(4kQ2)x/[(mv2)x - (4kQ2)] ]2 - x2 )

    With this equation, I found .003 nm. This was after multiplying the result with 109 so convert from meters to nanometers, of course.
  18. Mar 1, 2013 #17


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    OK, I have found my mistake...

    So KE=1.0248˙10-18

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