Question

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

Answer 1

speed of m/s in s

Answer 2

@satellite73

Answer 3

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

Answer 4

is used Ep=Ek is that correct?

Answer 5

Eelectron=Eproton? or am i overthinking it ?

Answer 6

Well you need the force

Answer 7

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

Answer 8

i think i am getting confused bettween force and freunecy

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

As in x and y directions

Answer 14

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

Answer 15

ok hold one i will get more probs

Answer 16

i know E=Ef-Ei

Answer 17

\[n_m - n_2~~~m=(3,4,5)\] 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.

Answer 18

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

Answer 19

the electrons are accelreated through a potential difference of 615 V and the spacing between the toms in the metal target is 0.243nm

Answer 20

sorry 0.234nm

Answer 21

@Astrophysics

Answer 22

@Michele_Laino

Answer 23

I agree with @Astrophysics

Answer 24

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

Answer 25

like from in glass vaccum

Answer 26

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

Answer 27

The angle of first order maximum is ?

Answer 28

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

Answer 29

where n=1

Answer 30

idk about braggg but we did the compton effect

Answer 31

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

Answer 32

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

Answer 33

ok

Answer 34

what to do next

Answer 35

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

Answer 36

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

Answer 37

^?

Answer 38

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

Answer 39

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

Answer 40

oh r u doing somthing like Ep=Ek

Answer 41

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

Answer 42

it is a simple substitution

Answer 43

ok

Answer 44

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?

Answer 45

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:
\[\Large E = eV = ...joules\]

Answer 46

2.188*10^-9

Answer 47

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

Answer 48

angle?

Answer 49

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:
\[\large p = \sqrt {2{m_e}E} = \sqrt {2 \times 9.11 \times {{10}^{ - 31}} \times 9.84 \times {{10}^{ - 17}}} = ...?\]

Answer 50

please complete, what do you get?

Answer 51

1.33*10^-23

Answer 52

coorrect!

Answer 53

correct!*

Answer 54

angle?

Answer 55

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

Answer 56

4.94*10^-11

Answer 57

that's right!

Answer 58

angle?

Answer 59

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

Answer 60

0.06049

Answer 61

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

Answer 62

6.05

Answer 63

I got this:
\[\Large \sin \theta = \frac{\lambda }{{2d}} = \frac{{4.94 \times {{10}^{ - 11}}}}{{2 \times 2.34 \times {{10}^{ - 10}}}} = 0.105555\]

Answer 64

i am in the end

Answer 65

from this the angle is 6.04 but the answer is 122

Answer 66

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

Answer 67

122 degrees?

Answer 68

please wait I check my computations

Answer 69

I got the same result

Answer 70

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