Calculating Maximum Resolution of Electron Microscope

[Airplane7]

Homework Statement

Electrons are accelerated by 2450 V in an electron microscope. What is the maximum possible resolution?
Me=9.11 x 10^-31kg and e= 1.60 x 10^-19 C

Homework Equations

I'm not so sure which equations to use for this question.. maybe the Planck's formula?
λ=h/p

The Attempt at a Solution

 

[rude man]

Hint #1: deBroglie
Hint #2: Rayleigh

 

[Airplane7]

Hint #1: deBroglie
Hint #2: Rayleigh

Thanks, Can you help me with the problem though? I'm confused..

 

[Airplane7]

So by using deBroglie equation lambda=h/p.. i'd have to do
6.63x10^-34Js / ((9.11x10^-31)(1.60x10^-19)) ??

 

[Airplane7]

when i find the solution, do i multiply the answer by 2450? to get the maximum resolution?

 

[vela]

So by using deBroglie equation lambda=h/p.. i'd have to do
6.63x10^-34Js / ((9.11x10^-31)(1.60x10^-19)) ??

No. Why would you multiply m and e? It doesn't make any sense.

 

[Redbelly98]

Moderator's note: I have moved this thread to Introductory Physics. In the future, please post questions of this level in Introductory Physics.

The information given for V can be used to find p.

 

[Airplane7]

No. Why would you multiply m and e? It doesn't make any sense.

Lol, I'm not sure.. I'm confused on solving the question..

So, to find p, it's p=mv right? so (9.11 x 10^-31kg)(2450 v) ?

When i find the solution of p, plug it into lambda=h/p, so (6.63x10^-34 Js) / p , will give me the maximum resolution?

 

[Airplane7]

Moderator's note: I have moved this thread to Introductory Physics. In the future, please post questions of this level in Introductory Physics.

The information given for V can be used to find p.

Thank you. Sorry, I'm new to the forum

 

[vela]

Lol, I'm not sure.. I'm confused on solving the question..

So, to find p, it's p=mv right? so (9.11 x 10^-31kg)(2450 v) ?

No. The v in p=mv represents the velocity of the particle. The V in 2450 V stands for volt; it's a unit of electric potential. They're not the same thing at all.

Read about electric potential in your textbook so you understand what the problem means when it says "Electrons are accelerated by 2450 V."

When i find the solution of p, plug it into lambda=h/p, so (6.63x10^-34 Js) / p , will give me the maximum resolution?

That's the basic idea, but you need to find the electron's momentum first.

 

[Airplane7]

Okay. I read up on it, but it didn't say much on "electrons accelerating"
I did see that 1 Volt = 1 J/C
So 2450 V = 2450 J/C
(1.60x10^-19 C)(2450 J/C) = ??

Man.. i am so confused

 

[Airplane7]

Okay. This is what I got, tell me if i did it correctly.

E= (1.60x10^19C)(2450J/C) = 3.92x10-16 J
V= Squareroot of √(2x3.92x10^-16) / (9.11x10^-31) = 2.93x10^7

So using de Broglie equation

lambda= (6.63x10^-34Js)/(9.11x10^-31)(2.93x10^7) = 2.48x10^-11 ?

 

[vela]

10^11? That's way off!

(Right method, though.)

 

[Airplane7]

Lol, aw man.. I thought I had it in the bag. What did I do wrong?

 

[Airplane7]

Wait.. I wrote, 10^-11, not 10^11. Am I still wrong?

 

[vela]

I see you edited your original answer so it says 10^-11 now instead of 10^11. That would be the right answer if you included the units.

 

[Airplane7]

Thank you for all your help!

Would the units be in meters?

 

[vela]

You tell me. :)

 

[Airplane7]

I guess so! :) Thanks for your help!