Competition - post a photo you've taken of nature >>

Physics equation

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!

Reply 1

Tgm11801

The voltage applied I assume

Reply 2

Kalabamboo

Original post by Tgm11801

The voltage applied I assume

Thanks! So U is the voltage applied? Is that right?

And also do you undertsand the equation I quoted above?

Reply 3

Kalabamboo

Anyone please? :frown:

Reply 4

homemadeclock

what don't you understand?

Reply 5

Kalabamboo

Original post by homemadeclock

what don't you understand?

Hey!

So
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.

Reply 6

homemadeclock

U is the voltage between the anode and cathode, are you familiar with E = V×Q?

Reply 7

username3442196

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

Reply 8

Kalabamboo

Original post by homemadeclock

U is the voltage between the anode and cathode, are you familiar with E = V×Q?

Thanks a lot! I wasn't familiar with it before but I am now that you have mentioned it! :biggrin:

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 dont quite understand what they are trying to explain here. I dont undretsand the content. What is this acceleration voltage and what causes it? Also, how do you go from

Attachment not found

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

(edited 5 years ago)

2.PNG4.8KB

Reply 9

Kalabamboo

Original post by old_teach

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

Thanks a lot! I wasn't familiar with it before but I am now that you have mentioned it! :biggrin:

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 dont quite understand what they are trying to explain here. What is this acceleration voltage and what causes it? Also, how do you go from

Attachment not found

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

Reply 10

username3442196

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)

Reply 11

Kalabamboo

Original post by old_teach

Thanks a lot! So I equated p=mv and p=sqrt(2mE) and rearranged and this gave me E = 1/2 mv^2. I also have to know that E = q x voltage but then how do you get equation for wavelength of electrons from there? Apologies again!

Reply 12

username3442196

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.

Reply 13

Kalabamboo

Original post by old_teach

Thank you very much! :smile:

I didn't trust them so I tried to calculate it myself haha - sorry! :wink:

just going to attach my workings in a few minutes - my laptop is constantly crashing for some reason Here it is :

Attachment not found

It didn't quite work out :/ to give 1.225/sqrtU

(edited 5 years ago)

Reply 14

Eimmanuel

Study Forum Helper

Original post by 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?

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

E = \dfrac{V}{d}

---Eqn(2)

Sub (2) into (1),

q \dfrac{V}{d} = ma

qV = mad ---Eqn(3)

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

\dfrac{1}{2}\times m (2ad) = \dfrac{1}{2}m v^2 = KE

where I have used v² = u² + 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:

\dfrac{1}{2}m v^2 = \dfrac{1}{2} \dfrac{m^2 v^2}{m} = \dfrac{p^2}{2m}

Original post by 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

\lambda = \dfrac{h}{\sqrt{2mE}} = \dfrac{h}{\sqrt{2m qV }}

h is known as the Planck constant, 6.63 × 10^-³⁴ m² kg / s

m is the mass of electron, 9.11 × 10^-³¹ kg

m is the electron charge, 1.60 × 10^-¹⁹ C

\lambda = \dfrac{ 6.63 \times 10^{-34}}{\sqrt{2 (9.11 \times 10^{-31}) (1.60 \times 10^{-19}) V }} = \dfrac{1.23 \times 10^{-9} \text{ m}}{\sqrt{V}}

Note that the V is the U in your note.

Reply 15

username3442196

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 :smile:

Reply 16

Kalabamboo

Original post by 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 :smile:

Ah yes thank you very much :smile:

Really appreciate your time!!

Reply 17

Kalabamboo

Original post by Eimmanuel

F = ma

qE = ma ---Eqn(1)

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

E = \dfrac{V}{d}

---Eqn(2)

Sub (2) into (1),

q \dfrac{V}{d} = ma

qV = mad ---Eqn(3)

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

\dfrac{1}{2}\times m (2ad) = \dfrac{1}{2}m v^2 = KE

where I have used v² = u² + 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:

\dfrac{1}{2}m v^2 = \dfrac{1}{2} \dfrac{m^2 v^2}{m} = \dfrac{p^2}{2m}

\lambda = \dfrac{h}{\sqrt{2mE}} = \dfrac{h}{\sqrt{2m qV }}

h is known as the Planck constant, 6.63 × 10^-³⁴ m² kg / s

m is the mass of electron, 9.11 × 10^-³¹ kg

m is the electron charge, 1.60 × 10^-¹⁹ C

\lambda = \dfrac{ 6.63 \times 10^{-34}}{\sqrt{2 (9.11 \times 10^{-31}) (1.60 \times 10^{-19}) V }} = \dfrac{1.23 \times 10^{-9} \text{ m}}{\sqrt{V}}

Note that the V is the U in your note.

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 :smile:

Reply 18

Eimmanuel

Study Forum Helper

Original post by 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 :smile:

No worry.