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

- anonymous

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- schrodinger

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- anonymous

speed of m/s in s

- anonymous

@satellite73

- Astrophysics

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

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## More answers

- anonymous

is used Ep=Ek is that correct?

- anonymous

Eelectron=Eproton? or am i overthinking it ?

- Astrophysics

Well you need the force

- anonymous

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

- anonymous

i think i am getting confused bettween force and freunecy

- Astrophysics

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

- 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

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

- anonymous

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

- Astrophysics

As in x and y directions

- Astrophysics

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

- anonymous

ok hold one i will get more probs

- anonymous

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- anonymous

i know E=Ef-Ei

- 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

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

- 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

sorry 0.234nm

- anonymous

@Astrophysics

- anonymous

@Michele_Laino

- Michele_Laino

I agree with @Astrophysics

- anonymous

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

- anonymous

like from in glass vaccum

- Michele_Laino

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

- anonymous

The angle of first order maximum is ?

- Michele_Laino

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

- Michele_Laino

where n=1

- anonymous

idk about braggg but we did the compton effect

- Michele_Laino

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

- Michele_Laino

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

- anonymous

ok

- anonymous

what to do next

- Michele_Laino

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

- Michele_Laino

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

- anonymous

^?

- Michele_Laino

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

- 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

oh r u doing somthing like Ep=Ek

- 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

it is a simple substitution

- anonymous

ok

- 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

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

2.188*10^-9

- Michele_Laino

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

- anonymous

angle?

- 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

please complete, what do you get?

- anonymous

1.33*10^-23

- Michele_Laino

coorrect!

- Michele_Laino

correct!*

- anonymous

angle?

- 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

4.94*10^-11

- Michele_Laino

that's right!

- anonymous

angle?

- 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

0.06049

- 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

6.05

- 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

i am in the end

- anonymous

from this the angle is 6.04 but the answer is 122

- Michele_Laino

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

- Michele_Laino

122 degrees?

- Michele_Laino

please wait I check my computations

- Michele_Laino

I got the same result

- 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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