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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 ?
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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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[F = ma = \frac{ mv }{ t }\] where the mass of an electron is \[\approx 9.11 \times 10^{31} kg\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is used Ep=Ek is that correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Eelectron=Eproton? or am i overthinking it ?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Well you need the force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can i do hf=1/2mv^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think i am getting confused bettween force and freunecy

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Yes, you are, what you're doing wouldn't make sense, that deals with the photoelectric effect.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2You 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
 one year ago
Best ResponseYou've already chosen the best response.2So for our case we have \[F = ma = \frac{ mv }{ t }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw do u know the component method when doing momentum questions?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2As in x and y directions

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2But yes, I know momentum, and how to deal with the problems.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok hold one i will get more probs

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[n_m  n_2~~~m=(3,4,5)\] so if you do \[n_3  n_1 = 3.41.51 = 1.89 eV\] if you do the others they would require an energy transfer above 2.0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in a glass vacuum electrons are accelatred trough a large potential difference

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the electrons are accelreated through a potential difference of 615 V and the spacing between the toms in the metal target is 0.243nm

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I agree with @Astrophysics

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i am done that question i am talkign about the new one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like from in glass vaccum

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I see only initial data, what quantity is requested to compute, please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The angle of first order maximum is ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to apply teh Bragg's formula: \[n\lambda = 2d\sin \theta \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0idk about braggg but we did the compton effect

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\lambda is given by the De Broglie relationship: \[\Large \lambda = \frac{h}{p}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the Compton effect is related to a collision between a photon and a free electron

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that your exercise is related to the diffraction by electrons

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the energy of the electrons, is: \[\Large E = \frac{{{p^2}}}{{2{m_e}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since our electron are classical (not relativistic) electrons. So we can write: \[\Large p = \sqrt {2{m_e}E} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please 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
 one year ago
Best ResponseYou've already chosen the best response.0oh r u doing somthing like Ep=Ek

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, 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
 one year ago
Best ResponseYou've already chosen the best response.1it is a simple substitution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, 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
 one year ago
Best ResponseYou've already chosen the best response.1sorry 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: \[\Large E = eV = ...joules\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got this: \[\Large E = eV = 1.6 \times {10^{  19}} \times 615 = 9.84 \times {10^{  17}}joules\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.1please complete, what do you get?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we can compute \lambda as follows: \[\Large \lambda = \frac{h}{p} = \frac{{6.62 \times {{10}^{  34}}}}{{1.34 \times {{10}^{  23}}}} = ...cm\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1finally, we can find your angle, as follows: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.9410  11}}{{2 \times 2.34 \times {{10}^{  8}}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry, our wavelength is measured in meters not in cm, so we have: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{  11}}}}{{2 \times 2.34 \times {{10}^{  10}}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got this: \[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{  11}}}}{{2 \times 2.34 \times {{10}^{  10}}}} = 0.105555\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from this the angle is 6.04 but the answer is 122

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we get: \[\theta = 6.06\;{\text{degrees}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please wait I check my computations

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got the same result

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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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