## 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 ?

1. anonymous

speed of m/s in s

2. anonymous

@satellite73

3. Astrophysics

$F = ma = \frac{ mv }{ t }$ where the mass of an electron is $\approx 9.11 \times 10^{-31} kg$

4. anonymous

is used Ep=Ek is that correct?

5. anonymous

Eelectron=Eproton? or am i overthinking it ?

6. Astrophysics

Well you need the force

7. anonymous

can i do hf=1/2mv^2?

8. anonymous

i think i am getting confused bettween force and freunecy

9. Astrophysics

Yes, you are, what you're doing wouldn't make sense, that deals with the photoelectric effect.

10. Astrophysics

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

So for our case we have $F = ma = \frac{ mv }{ t }$

12. anonymous

btw do u know the component method when doing momentum questions?

13. Astrophysics

As in x and y directions

14. Astrophysics

But yes, I know momentum, and how to deal with the problems.

15. anonymous

ok hold one i will get more probs

16. anonymous

17. anonymous

i know E=Ef-Ei

18. Astrophysics

$n_m - n_2~~~m=(3,4,5)$ 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

in a glass vacuum electrons are accelatred trough a large potential difference

20. anonymous

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

sorry 0.234nm

22. anonymous

@Astrophysics

23. anonymous

@Michele_Laino

24. Michele_Laino

I agree with @Astrophysics

25. anonymous

no i am done that question i am talkign about the new one

26. anonymous

like from in glass vaccum

27. Michele_Laino

I see only initial data, what quantity is requested to compute, please?

28. anonymous

The angle of first order maximum is ?

29. Michele_Laino

we have to apply teh Bragg's formula: $n\lambda = 2d\sin \theta$

30. Michele_Laino

where n=1

31. anonymous

idk about braggg but we did the compton effect

32. Michele_Laino

\lambda is given by the De Broglie relationship: $\Large \lambda = \frac{h}{p}$

33. Michele_Laino

the Compton effect is related to a collision between a photon and a free electron

34. anonymous

ok

35. anonymous

what to do next

36. Michele_Laino

I think that your exercise is related to the diffraction by electrons

37. Michele_Laino

the energy of the electrons, is: $\Large E = \frac{{{p^2}}}{{2{m_e}}}$

38. anonymous

^?

39. Michele_Laino

since our electron are classical (not relativistic) electrons. So we can write: $\Large p = \sqrt {2{m_e}E}$

40. Michele_Laino

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

oh r u doing somthing like Ep=Ek

42. Michele_Laino

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

it is a simple substitution

44. anonymous

ok

45. Michele_Laino

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

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

2.188*10^-9

48. Michele_Laino

I got this: $\Large E = eV = 1.6 \times {10^{ - 19}} \times 615 = 9.84 \times {10^{ - 17}}joules$

49. anonymous

angle?

50. Michele_Laino

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

please complete, what do you get?

52. anonymous

1.33*10^-23

53. Michele_Laino

coorrect!

54. Michele_Laino

correct!*

55. anonymous

angle?

56. Michele_Laino

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

4.94*10^-11

58. Michele_Laino

that's right!

59. anonymous

angle?

60. Michele_Laino

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

0.06049

62. Michele_Laino

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

6.05

64. Michele_Laino

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

i am in the end

66. anonymous

from this the angle is 6.04 but the answer is 122

67. Michele_Laino

yes! we get: $\theta = 6.06\;{\text{degrees}}$

68. Michele_Laino

122 degrees?

69. Michele_Laino

please wait I check my computations

70. Michele_Laino

I got the same result

71. Michele_Laino

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: $\lambda {\text{ = 0}}{\text{.494}}\;{\text{Angstrom}}$