## Sapling Homework #22

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Racquel Fox 3L
Posts: 40
Joined: Wed Sep 30, 2020 10:11 pm

### Sapling Homework #22

You can use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0339 nm. What is the kinetic energy of an electron in the beam, expressed in electron volts?

I first approached this problem by trying to find the frequency from deriving the c=lambda*v equation, but that got the wrong answer and Sapling said to use the lambda=h/(m*v) equation instead. I'm wondering why I would use that equation to find the frequency and not the v=c/lambda?

Stephen Min 2K
Posts: 40
Joined: Wed Sep 30, 2020 10:01 pm

### Re: Sapling Homework #22

The v in de Broglie's equation and the kinetic energy equation stands for velocity, not frequency. So, using c=lambda*v would not allow us to solve for kinetic energy, which requires velocity. Also, I believe de Broglie's equation would be used because we are dealing with an electron.

Angel Gutierrez 2E
Posts: 43
Joined: Wed Sep 30, 2020 9:49 pm

### Re: Sapling Homework #22

you would use de brogile equation in order to find the kinetic energy which is 1/2mv^2. when rearragning the debrogile equation, you can solve for v to plug v back into the kinetic energy equation!

Ashley Kim 3I
Posts: 45
Joined: Wed Sep 30, 2020 9:34 pm
Been upvoted: 1 time

### Re: Sapling Homework #22

Hi! For this question, you would first use DeBroglie's equation ( v = h/m*lambda) to find the velocity of the electron with the given wavelength and the mass of an electron (9.11 x 10^-31kg). After finding the velocity, you can plug the value into the kinetic energy equation 1/2mv^2.