anonymous
  • anonymous
The electron are accelrated to a speed of 2.40*10^7 in 1.8*10^-9, the force experinced by an electron ?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
speed of m/s in s
anonymous
  • anonymous
@satellite73
Astrophysics
  • Astrophysics
\[F = ma = \frac{ mv }{ t }\] where the mass of an electron is \[\approx 9.11 \times 10^{-31} kg\]

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anonymous
  • anonymous
is used Ep=Ek is that correct?
anonymous
  • anonymous
Eelectron=Eproton? or am i overthinking it ?
Astrophysics
  • Astrophysics
Well you need the force
anonymous
  • anonymous
can i do hf=1/2mv^2?
anonymous
  • anonymous
i think i am getting confused bettween force and freunecy
Astrophysics
  • Astrophysics
Yes, you are, what you're doing wouldn't make sense, that deals with the photoelectric effect.
Astrophysics
  • Astrophysics
You want the force F, so we use Newton's second law of motion \[F = ma\] where acceleration is \[a = \frac{ \Delta v }{ \Delta t }\]
Astrophysics
  • Astrophysics
So for our case we have \[F = ma = \frac{ mv }{ t }\]
anonymous
  • anonymous
btw do u know the component method when doing momentum questions?
Astrophysics
  • Astrophysics
As in x and y directions
Astrophysics
  • Astrophysics
But yes, I know momentum, and how to deal with the problems.
anonymous
  • anonymous
ok hold one i will get more probs
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
i know E=Ef-Ei
Astrophysics
  • 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.
anonymous
  • anonymous
in a glass vacuum electrons are accelatred trough a large potential difference
anonymous
  • 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
anonymous
  • anonymous
sorry 0.234nm
anonymous
  • anonymous
@Astrophysics
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
I agree with @Astrophysics
anonymous
  • anonymous
no i am done that question i am talkign about the new one
anonymous
  • anonymous
like from in glass vaccum
Michele_Laino
  • Michele_Laino
I see only initial data, what quantity is requested to compute, please?
anonymous
  • anonymous
The angle of first order maximum is ?
Michele_Laino
  • Michele_Laino
we have to apply teh Bragg's formula: \[n\lambda = 2d\sin \theta \]
Michele_Laino
  • Michele_Laino
where n=1
anonymous
  • anonymous
idk about braggg but we did the compton effect
Michele_Laino
  • Michele_Laino
\lambda is given by the De Broglie relationship: \[\Large \lambda = \frac{h}{p}\]
Michele_Laino
  • Michele_Laino
the Compton effect is related to a collision between a photon and a free electron
anonymous
  • anonymous
ok
anonymous
  • anonymous
what to do next
Michele_Laino
  • Michele_Laino
I think that your exercise is related to the diffraction by electrons
Michele_Laino
  • Michele_Laino
the energy of the electrons, is: \[\Large E = \frac{{{p^2}}}{{2{m_e}}}\]
anonymous
  • anonymous
^?
Michele_Laino
  • Michele_Laino
since our electron are classical (not relativistic) electrons. So we can write: \[\Large p = \sqrt {2{m_e}E} \]
Michele_Laino
  • Michele_Laino
please note that the energy of our electron is all kinetic energy: \[\Large E = \frac{1}{2}{m_e}{v^2} = \frac{1}{2}{m_e}{\left( {\frac{p}{{{m_e}}}} \right)^2} = \frac{{{p^2}}}{{2{m_e}}}\]
anonymous
  • anonymous
oh r u doing somthing like Ep=Ek
Michele_Laino
  • Michele_Laino
no, I have changed variable, I expressed the kinetic energy of electron as a function of its momentum: \[\Large {v_e} = \frac{p}{{{m_e}}}\]
Michele_Laino
  • Michele_Laino
it is a simple substitution
anonymous
  • anonymous
ok
Michele_Laino
  • 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?
Michele_Laino
  • 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: \[\Large E = eV = ...joules\]
anonymous
  • anonymous
2.188*10^-9
Michele_Laino
  • Michele_Laino
I got this: \[\Large E = eV = 1.6 \times {10^{ - 19}} \times 615 = 9.84 \times {10^{ - 17}}joules\]
anonymous
  • anonymous
angle?
Michele_Laino
  • 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: \[\large p = \sqrt {2{m_e}E} = \sqrt {2 \times 9.11 \times {{10}^{ - 31}} \times 9.84 \times {{10}^{ - 17}}} = ...?\]
Michele_Laino
  • Michele_Laino
please complete, what do you get?
anonymous
  • anonymous
1.33*10^-23
Michele_Laino
  • Michele_Laino
coorrect!
Michele_Laino
  • Michele_Laino
correct!*
anonymous
  • anonymous
angle?
Michele_Laino
  • Michele_Laino
now we can compute \lambda as follows: \[\Large \lambda = \frac{h}{p} = \frac{{6.62 \times {{10}^{ - 34}}}}{{1.34 \times {{10}^{ - 23}}}} = ...cm\]
anonymous
  • anonymous
4.94*10^-11
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
angle?
Michele_Laino
  • Michele_Laino
finally, we can find your angle, as follows: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.9410 - 11}}{{2 \times 2.34 \times {{10}^{ - 8}}}} = ...\]
anonymous
  • anonymous
0.06049
Michele_Laino
  • Michele_Laino
sorry, our wavelength is measured in meters not in cm, so we have: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = ...\]
anonymous
  • anonymous
6.05
Michele_Laino
  • Michele_Laino
I got this: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = 0.105555\]
anonymous
  • anonymous
i am in the end
anonymous
  • anonymous
from this the angle is 6.04 but the answer is 122
Michele_Laino
  • Michele_Laino
yes! we get: \[\theta = 6.06\;{\text{degrees}}\]
Michele_Laino
  • Michele_Laino
122 degrees?
Michele_Laino
  • Michele_Laino
please wait I check my computations
Michele_Laino
  • Michele_Laino
I got the same result
Michele_Laino
  • 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: \[\lambda {\text{ = 0}}{\text{.494}}\;{\text{Angstrom}}\]

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