Recent content by Airplane7

I guess so! :) Thanks for your help!

Thank you for all your help! Would the units be in meters?

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

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

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 ?

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

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

Sorry, I'm new to the forum..

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?

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

Can someone give me a hint on how to figure out the solution?

Homework Statement The solution of the Schrodinger equation for atom depends on four quantum numbers: the principal n, the orbital l, magnetic m1, and the spin Ms n = 1, 2, 3, 4, ... (integers) l = 0, 1, 2, ... (n-1) m1 = -1, -1+1,...0...1-1, 1 Ms = -1/2, 1/2 List all possible values of...

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

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

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...