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

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SUMMARY

The kinetic energy of an electron with a wavelength of 0.850 x 10-10 m can be calculated using the de Broglie wavelength concept. The relevant equation is derived from the relationship between wavelength and momentum, given by λ = h/p, where h is Planck's constant and p is momentum. The kinetic energy can then be expressed as K = p²/2m, where m is the mass of the electron. Thus, the correct approach involves using the de Broglie wavelength to find the momentum and subsequently the kinetic energy.

PREREQUISITES
  • Understanding of de Broglie wavelength
  • Familiarity with Planck's constant (h)
  • Knowledge of kinetic energy formula (K = p²/2m)
  • Basic principles of quantum mechanics
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  • Research the de Broglie wavelength and its applications in quantum mechanics
  • Learn how to calculate momentum from wavelength using p = h/λ
  • Study the relationship between kinetic energy and momentum in quantum particles
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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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