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anonymous

  • one year ago

The electron are accelrated to a speed of 2.40*10^7 in 1.8*10^-9, the force experinced by an electron ?

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  1. anonymous
    • one year ago
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    speed of m/s in s

  2. anonymous
    • one year ago
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    @satellite73

  3. Astrophysics
    • one year ago
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    \[F = ma = \frac{ mv }{ t }\] where the mass of an electron is \[\approx 9.11 \times 10^{-31} kg\]

  4. anonymous
    • one year ago
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    is used Ep=Ek is that correct?

  5. anonymous
    • one year ago
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    Eelectron=Eproton? or am i overthinking it ?

  6. Astrophysics
    • one year ago
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    Well you need the force

  7. anonymous
    • one year ago
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    can i do hf=1/2mv^2?

  8. anonymous
    • one year ago
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    i think i am getting confused bettween force and freunecy

  9. Astrophysics
    • one year ago
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    Yes, you are, what you're doing wouldn't make sense, that deals with the photoelectric effect.

  10. Astrophysics
    • one year ago
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    You want the force F, so we use Newton's second law of motion \[F = ma\] where acceleration is \[a = \frac{ \Delta v }{ \Delta t }\]

  11. Astrophysics
    • one year ago
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    So for our case we have \[F = ma = \frac{ mv }{ t }\]

  12. anonymous
    • one year ago
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    btw do u know the component method when doing momentum questions?

  13. Astrophysics
    • one year ago
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    As in x and y directions

  14. Astrophysics
    • one year ago
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    But yes, I know momentum, and how to deal with the problems.

  15. anonymous
    • one year ago
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    ok hold one i will get more probs

  16. anonymous
    • one year ago
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  17. anonymous
    • one year ago
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    i know E=Ef-Ei

  18. Astrophysics
    • one year ago
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    \[n_m - n_2~~~m=(3,4,5)\] so if you do \[n_3 - n_1 = 3.4-1.51 = 1.89 eV\] if you do the others they would require an energy transfer above 2.0.

  19. anonymous
    • one year ago
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    in a glass vacuum electrons are accelatred trough a large potential difference

  20. anonymous
    • one year ago
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    the electrons are accelreated through a potential difference of 615 V and the spacing between the toms in the metal target is 0.243nm

  21. anonymous
    • one year ago
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    sorry 0.234nm

  22. anonymous
    • one year ago
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    @Astrophysics

  23. anonymous
    • one year ago
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    @Michele_Laino

  24. Michele_Laino
    • one year ago
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    I agree with @Astrophysics

  25. anonymous
    • one year ago
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    no i am done that question i am talkign about the new one

  26. anonymous
    • one year ago
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    like from in glass vaccum

  27. Michele_Laino
    • one year ago
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    I see only initial data, what quantity is requested to compute, please?

  28. anonymous
    • one year ago
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    The angle of first order maximum is ?

  29. Michele_Laino
    • one year ago
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    we have to apply teh Bragg's formula: \[n\lambda = 2d\sin \theta \]

  30. Michele_Laino
    • one year ago
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    where n=1

  31. anonymous
    • one year ago
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    idk about braggg but we did the compton effect

  32. Michele_Laino
    • one year ago
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    \lambda is given by the De Broglie relationship: \[\Large \lambda = \frac{h}{p}\]

  33. Michele_Laino
    • one year ago
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    the Compton effect is related to a collision between a photon and a free electron

  34. anonymous
    • one year ago
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    ok

  35. anonymous
    • one year ago
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    what to do next

  36. Michele_Laino
    • one year ago
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    I think that your exercise is related to the diffraction by electrons

  37. Michele_Laino
    • one year ago
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    the energy of the electrons, is: \[\Large E = \frac{{{p^2}}}{{2{m_e}}}\]

  38. anonymous
    • one year ago
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    ^?

  39. Michele_Laino
    • one year ago
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    since our electron are classical (not relativistic) electrons. So we can write: \[\Large p = \sqrt {2{m_e}E} \]

  40. Michele_Laino
    • one year ago
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    please note that the energy of our electron is all kinetic energy: \[\Large E = \frac{1}{2}{m_e}{v^2} = \frac{1}{2}{m_e}{\left( {\frac{p}{{{m_e}}}} \right)^2} = \frac{{{p^2}}}{{2{m_e}}}\]

  41. anonymous
    • one year ago
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    oh r u doing somthing like Ep=Ek

  42. Michele_Laino
    • one year ago
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    no, I have changed variable, I expressed the kinetic energy of electron as a function of its momentum: \[\Large {v_e} = \frac{p}{{{m_e}}}\]

  43. Michele_Laino
    • one year ago
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    it is a simple substitution

  44. anonymous
    • one year ago
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    ok

  45. Michele_Laino
    • one year ago
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    now, we have to compute the momentum of our electron, as follows: \[\Large p = \sqrt {2{m_e}E} \] with your initial data. What do you get?

  46. Michele_Laino
    • one year ago
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    sorry after the acceleration, the kinetic energy gained by our electron is coming from the potential energy due to the electric potential, so we can write: \[\Large E = eV = ...joules\]

  47. anonymous
    • one year ago
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    2.188*10^-9

  48. Michele_Laino
    • one year ago
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    I got this: \[\Large E = eV = 1.6 \times {10^{ - 19}} \times 615 = 9.84 \times {10^{ - 17}}joules\]

  49. anonymous
    • one year ago
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    angle?

  50. Michele_Laino
    • one year ago
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    we have to compute the wavelength associated to our electron first, in order to do that, we have to compute its momentum, so we have to compute this: \[\large p = \sqrt {2{m_e}E} = \sqrt {2 \times 9.11 \times {{10}^{ - 31}} \times 9.84 \times {{10}^{ - 17}}} = ...?\]

  51. Michele_Laino
    • one year ago
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    please complete, what do you get?

  52. anonymous
    • one year ago
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    1.33*10^-23

  53. Michele_Laino
    • one year ago
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    coorrect!

  54. Michele_Laino
    • one year ago
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    correct!*

  55. anonymous
    • one year ago
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    angle?

  56. Michele_Laino
    • one year ago
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    now we can compute \lambda as follows: \[\Large \lambda = \frac{h}{p} = \frac{{6.62 \times {{10}^{ - 34}}}}{{1.34 \times {{10}^{ - 23}}}} = ...cm\]

  57. anonymous
    • one year ago
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    4.94*10^-11

  58. Michele_Laino
    • one year ago
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    that's right!

  59. anonymous
    • one year ago
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    angle?

  60. Michele_Laino
    • one year ago
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    finally, we can find your angle, as follows: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.9410 - 11}}{{2 \times 2.34 \times {{10}^{ - 8}}}} = ...\]

  61. anonymous
    • one year ago
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    0.06049

  62. Michele_Laino
    • one year ago
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    sorry, our wavelength is measured in meters not in cm, so we have: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = ...\]

  63. anonymous
    • one year ago
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    6.05

  64. Michele_Laino
    • one year ago
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    I got this: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = 0.105555\]

  65. anonymous
    • one year ago
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    i am in the end

  66. anonymous
    • one year ago
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    from this the angle is 6.04 but the answer is 122

  67. Michele_Laino
    • one year ago
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    yes! we get: \[\theta = 6.06\;{\text{degrees}}\]

  68. Michele_Laino
    • one year ago
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    122 degrees?

  69. Michele_Laino
    • one year ago
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    please wait I check my computations

  70. Michele_Laino
    • one year ago
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    I got the same result

  71. Michele_Laino
    • one year ago
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    I made the same computation using a slight different procedure, and I got the same result again, in particular, the wavelength of your electrons is: \[\lambda {\text{ = 0}}{\text{.494}}\;{\text{Angstrom}}\]

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