# Physics equation

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Hi, so what is U here? I think qe is charge of electron. I am not sure what the equation E =qe.U is for? Dont really get that.

Help please!

Help please!

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(Original post by

The voltage applied I assume

**Tgm11801**)The voltage applied I assume

And also do you undertsand the equation I quoted above?

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(Original post by

what don't you understand?

**homemadeclock**)what don't you understand?

so what is U here? I think qe is charge of electron. I am not sure what the equation E =qe.U is for? Dont really get that.

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(Original post by

U is the voltage between the anode and cathode, are you familiar with

**homemadeclock**)U is the voltage between the anode and cathode, are you familiar with

**E = V×Q?**
Last edited by Kalabamboo; 2 years ago

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(Original post by

E is energy. Equation is 'energy = charge x voltage'

**old_teach**)E is energy. Equation is 'energy = charge x voltage'

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#11

This may help ...

The equation for wavelength is = h / momentum (p).

Momentum depends on the ke, E:

since E = 1/2 m v2, p = mv = sqrt(2mE)

since E = q x voltage, they get an equation for wavelength (of electrons - this may be an odd concept if you've not done much physics!) depending on accelerating voltage (which they call U)

I guess being able to change the wavelength of electrons is useful for electron microscopes.

(btw they are giving the electrons ke by accelerating them towards a large +ve voltage)

The equation for wavelength is = h / momentum (p).

Momentum depends on the ke, E:

since E = 1/2 m v2, p = mv = sqrt(2mE)

since E = q x voltage, they get an equation for wavelength (of electrons - this may be an odd concept if you've not done much physics!) depending on accelerating voltage (which they call U)

I guess being able to change the wavelength of electrons is useful for electron microscopes.

(btw they are giving the electrons ke by accelerating them towards a large +ve voltage)

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(Original post by

This may help ...

The equation for wavelength is = h / momentum (p).

Momentum depends on the ke, E:

since E = 1/2 m v2, p = mv = sqrt(2mE)

since E = q x voltage, they get an equation for wavelength (of electrons - this may be an odd concept if you've not done much physics!) depending on accelerating voltage (which they call U)

I guess being able to change the wavelength of electrons is useful for electron microscopes.

(btw they are giving the electrons ke by accelerating them towards a large +ve voltage)

**old_teach**)This may help ...

The equation for wavelength is = h / momentum (p).

Momentum depends on the ke, E:

since E = 1/2 m v2, p = mv = sqrt(2mE)

since E = q x voltage, they get an equation for wavelength (of electrons - this may be an odd concept if you've not done much physics!) depending on accelerating voltage (which they call U)

I guess being able to change the wavelength of electrons is useful for electron microscopes.

(btw they are giving the electrons ke by accelerating them towards a large +ve voltage)

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#13

wavelength = h / p (de Broglie's equation - just accept it's true!)

= h / (mv)

= h / sqrt(2m E) (remember we know E = 1/2 m v2)

= h / sqrt(2m qU) (E = qU)

= numbers / sqrt(U)

where numbers = h / sqrt(2mq). I trust them = 1.225 (x 10-9)

hope that's ok now.

= h / (mv)

= h / sqrt(2m E) (remember we know E = 1/2 m v2)

= h / sqrt(2m qU) (E = qU)

= numbers / sqrt(U)

where numbers = h / sqrt(2mq). I trust them = 1.225 (x 10-9)

hope that's ok now.

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(Original post by

wavelength = h / p (de Broglie's equation - just accept it's true!)

= h / (mv)

= h / sqrt(2m E) (remember we know E = 1/2 m v2)

= h / sqrt(2m qU) (E = qU)

= numbers / sqrt(U)

where numbers = h / sqrt(2mq). I trust them = 1.225 (x 10-9)

hope that's ok now.

**old_teach**)wavelength = h / p (de Broglie's equation - just accept it's true!)

= h / (mv)

= h / sqrt(2m E) (remember we know E = 1/2 m v2)

= h / sqrt(2m qU) (E = qU)

= numbers / sqrt(U)

where numbers = h / sqrt(2mq). I trust them = 1.225 (x 10-9)

hope that's ok now.

Last edited by Kalabamboo; 2 years ago

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#15

(Original post by

Thanks a lot! I wasn't familiar with it before but I am now that you have mentioned it! So how is the energy of electrons used for imaging connected to magnitude of the acceleration voltage between the anode and cathode of electron beam source. So I know it simply state in the text the relationship between these two but I don’t quite understand what they are trying to explain here. I don’t understand the content. What is this acceleration voltage and what causes it?

**Kalabamboo**)Thanks a lot! I wasn't familiar with it before but I am now that you have mentioned it! So how is the energy of electrons used for imaging connected to magnitude of the acceleration voltage between the anode and cathode of electron beam source. So I know it simply state in the text the relationship between these two but I don’t quite understand what they are trying to explain here. I don’t understand the content. What is this acceleration voltage and what causes it?

Assume that there is a pair of metallic parallel plates separated by a distance

*d*and is connected to a potential difference of say

*V*. A uniform electric field

*E*is set up between the pair of metallic parallel plates. When a stationary electron

*q*is placed in just outside the negative potential plate, it is subjected to an electric force

*qE*. So it would accelerate toward the positive potential plate.

From Newton’s 2

^{nd}law,

*F*=

*ma*

*qE*=

*ma*---Eqn(1)

The uniform electric field between the pair of metallic parallel plates is “defined” by

Sub (2) into (1),

*qV*=

*mad*---Eqn(3)

If you manipulate the left-hand side of (3) by

where I have used v

^{2}= u

^{2}+ 2as.

So we can say that the gain in KE of the electron is equal to

*qV*.

Note that we can KE in terms of momentum of the electron:

(Original post by

Also, how do you go from to I really apologise for my dumb questions! I have done mechanics As level and got an A in that but other than that I am not very familiar with physics :/ - this is biophysics stuff for med school

**Kalabamboo**)Also, how do you go from to I really apologise for my dumb questions! I have done mechanics As level and got an A in that but other than that I am not very familiar with physics :/ - this is biophysics stuff for med school

*h*is known as the Planck constant, 6.63 × 10

^{-}

^{34}m

^{2}kg / s

*m*is the mass of electron, 9.11 × 10

^{-}

^{31}kg

*m*is the electron charge, 1.60 × 10

^{-}

^{19}C

Note that the

*V*is the

*U*in your note.

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#16

charge on electron is not -1!

ignore the sign - doesn't work well under sqrt! The equation has assumed electron is negative.

q = 1.6 x 10-19 C.

It worked for me

ignore the sign - doesn't work well under sqrt! The equation has assumed electron is negative.

q = 1.6 x 10-19 C.

It worked for me

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(Original post by

charge on electron is not -1!

ignore the sign - doesn't work well under sqrt! The equation has assumed electron is negative.

q = 1.6 x 10-19 C.

It worked for me

**old_teach**)charge on electron is not -1!

ignore the sign - doesn't work well under sqrt! The equation has assumed electron is negative.

q = 1.6 x 10-19 C.

It worked for me

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reply

(Original post by

You need some basic concepts of electric field.

Assume that there is a pair of metallic parallel plates separated by a distance

From Newton’s 2

The uniform electric field between the pair of metallic parallel plates is “defined” by

Sub (2) into (1),

If you manipulate the left-hand side of (3) by

where I have used v

So we can say that the gain in KE of the electron is equal to

Note that we can KE in terms of momentum of the electron:

**Eimmanuel**)You need some basic concepts of electric field.

Assume that there is a pair of metallic parallel plates separated by a distance

*d*and is connected to a potential difference of say*V*. A uniform electric field*E*is set up between the pair of metallic parallel plates. When a stationary electron*q*is placed in just outside the negative potential plate, it is subjected to an electric force*qE*. So it would accelerate toward the positive potential plate.From Newton’s 2

^{nd}law,*F*=

*ma*

*qE*=

*ma*---Eqn(1)

The uniform electric field between the pair of metallic parallel plates is “defined” by

Sub (2) into (1),

*qV*=

*mad*---Eqn(3)

If you manipulate the left-hand side of (3) by

where I have used v

^{2}= u^{2}+ 2as.So we can say that the gain in KE of the electron is equal to

*qV*.Note that we can KE in terms of momentum of the electron:

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#19

(Original post by

Thank you so much! My brain is in sleep mode at the moment haha but I will definitely reply back to this and also to your kind reply back to my other question! Really appreciate your help

**Kalabamboo**)Thank you so much! My brain is in sleep mode at the moment haha but I will definitely reply back to this and also to your kind reply back to my other question! Really appreciate your help

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