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

Physics
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speed of m/s in s
\[F = ma = \frac{ mv }{ t }\] where the mass of an electron is \[\approx 9.11 \times 10^{-31} kg\]

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is used Ep=Ek is that correct?
Eelectron=Eproton? or am i overthinking it ?
Well you need the force
can i do hf=1/2mv^2?
i think i am getting confused bettween force and freunecy
Yes, you are, what you're doing wouldn't make sense, that deals with the photoelectric effect.
You want the force F, so we use Newton's second law of motion \[F = ma\] where acceleration is \[a = \frac{ \Delta v }{ \Delta t }\]
So for our case we have \[F = ma = \frac{ mv }{ t }\]
btw do u know the component method when doing momentum questions?
As in x and y directions
But yes, I know momentum, and how to deal with the problems.
ok hold one i will get more probs
1 Attachment
i know E=Ef-Ei
\[n_m - n_2~~~m=(3,4,5)\] 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.
in a glass vacuum electrons are accelatred trough a large potential difference
the electrons are accelreated through a potential difference of 615 V and the spacing between the toms in the metal target is 0.243nm
sorry 0.234nm
I agree with @Astrophysics
no i am done that question i am talkign about the new one
like from in glass vaccum
I see only initial data, what quantity is requested to compute, please?
The angle of first order maximum is ?
we have to apply teh Bragg's formula: \[n\lambda = 2d\sin \theta \]
where n=1
idk about braggg but we did the compton effect
\lambda is given by the De Broglie relationship: \[\Large \lambda = \frac{h}{p}\]
the Compton effect is related to a collision between a photon and a free electron
ok
what to do next
I think that your exercise is related to the diffraction by electrons
the energy of the electrons, is: \[\Large E = \frac{{{p^2}}}{{2{m_e}}}\]
^?
since our electron are classical (not relativistic) electrons. So we can write: \[\Large p = \sqrt {2{m_e}E} \]
please note that the energy of our electron is all kinetic energy: \[\Large E = \frac{1}{2}{m_e}{v^2} = \frac{1}{2}{m_e}{\left( {\frac{p}{{{m_e}}}} \right)^2} = \frac{{{p^2}}}{{2{m_e}}}\]
oh r u doing somthing like Ep=Ek
no, I have changed variable, I expressed the kinetic energy of electron as a function of its momentum: \[\Large {v_e} = \frac{p}{{{m_e}}}\]
it is a simple substitution
ok
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?
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: \[\Large E = eV = ...joules\]
2.188*10^-9
I got this: \[\Large E = eV = 1.6 \times {10^{ - 19}} \times 615 = 9.84 \times {10^{ - 17}}joules\]
angle?
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: \[\large p = \sqrt {2{m_e}E} = \sqrt {2 \times 9.11 \times {{10}^{ - 31}} \times 9.84 \times {{10}^{ - 17}}} = ...?\]
please complete, what do you get?
1.33*10^-23
coorrect!
correct!*
angle?
now we can compute \lambda as follows: \[\Large \lambda = \frac{h}{p} = \frac{{6.62 \times {{10}^{ - 34}}}}{{1.34 \times {{10}^{ - 23}}}} = ...cm\]
4.94*10^-11
that's right!
angle?
finally, we can find your angle, as follows: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.9410 - 11}}{{2 \times 2.34 \times {{10}^{ - 8}}}} = ...\]
0.06049
sorry, our wavelength is measured in meters not in cm, so we have: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = ...\]
6.05
I got this: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = 0.105555\]
i am in the end
from this the angle is 6.04 but the answer is 122
yes! we get: \[\theta = 6.06\;{\text{degrees}}\]
122 degrees?
please wait I check my computations
I got the same result
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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