What is the kinetic energy of an electron with a wavelength of 0.850 x 10^-10 m?

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of an electron given its wavelength of 0.850 x 10^-10 m, touching on concepts related to wave-particle duality and kinetic energy equations.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants express uncertainty about the appropriate equations to use for relating wavelength to kinetic energy. Some suggest looking for relevant equations in textbooks, while others mention specific formulas like E = hc/λ and K = 1/2MV², questioning their applicability.

Discussion Status

There is an ongoing exploration of relevant equations, with some participants indicating a need to revisit foundational concepts such as the de Broglie wavelength. Guidance has been offered regarding the relevance of specific equations, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are grappling with the definitions and applications of kinetic energy and wavelength in the context of quantum mechanics, indicating a possible gap in foundational understanding.

Cowtipper
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Homework Statement


If an electron has a measured wavelength of 0.850 x 10^-10 m, what is its kinetic energy?


Homework Equations


I'm not sure.


The Attempt at a Solution


And once again, I'm not too sure. Where do I start?
 
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Find an equation in your textbook that deals with the wavelength of electrons or particles in general. That would be a start.
 
Well, I've got this:

I'm not sure if it's one to use.

E = hc/\lambda

Similarly, I have this one:

K=1/2MV^{2}

However, I'm not sure where or how to use it in this instance, or if I have to use it at all.

Man, this stuff is getting tough. I was doing well there for a while too...
 
Cowtipper said:
Well, I've got this:

I'm not sure if it's one to use.

E = hc/\lambda

No, this equation will not work. Look back in your notes and try to figure out why.

Look up "de Broglie wavelength."
 
hage567 said:
No, this equation will not work. Look back in your notes and try to figure out why.

Look up "de Broglie wavelength."

Aha! That is all I needed to know.

Thanks a lot!
 

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