How to Calculate the de Broglie Wavelength of a 5.0 eV Electron?

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SUMMARY

The de Broglie wavelength of a 5.0 eV electron can be calculated using the formula λ = h/p, where p is the momentum derived from the kinetic energy equation E = p²/(2m). The correct momentum for a 5.0 eV electron is calculated as p = √(2mE), resulting in a wavelength of approximately 5.48 x 10-10 m after converting energy from electron volts to joules. A common mistake is neglecting to include the factor of 2 in the kinetic energy formula and failing to convert eV to joules, which leads to incorrect wavelength calculations.

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  • Understanding of quantum mechanics concepts, specifically de Broglie wavelength.
  • Familiarity with the relationship between energy, momentum, and mass.
  • Knowledge of unit conversions, particularly between electron volts and joules.
  • Basic proficiency in algebra and solving equations.
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  • Study the derivation of the de Broglie wavelength formula.
  • Learn about the conversion of electron volts to joules and its significance in calculations.
  • Explore kinetic energy equations for particles at relativistic speeds.
  • Investigate the implications of potential difference on electron acceleration and energy.
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Students and professionals in physics, particularly those studying quantum mechanics, as well as educators looking to clarify concepts related to electron behavior and wave-particle duality.

StudentofPhysics
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1. What is the wavelength of a 5.0 eV photon?
What is the de Broglie wavelength of a 5.0 eV electron?




2.E = hc/lambda
lambda = h/momentum




3. I know the first one simply 248 nm for the wavelngth.

I don't know how to find the de Broglie though since I don't know what the momentum of the electron is.
 
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Try finding it form:
E=\frac{p^2}{2m}
 
ok i tried it with that formula this way:

5 eV = p^2 / (9.11 x 10^-31kg)
p= 6.75 x 10^-17

this gave me a wavelength of 9.82 x 10^-18 m and wasn't correct...


I'm still doing something wrong
 
I'm having the same problem with this question:
An electron, starting from rest, accelerates through a potential difference of 415 V. What is the final de Broglie wavelength of the electron, assuming that its final speed is much less than the speed of light?

Applying that formula i get 3.41e-20

For both of these questions I figured maybe it was in nm, so i switched to meters by adding ^-9 and they were still wrong...
 
StudentofPhysics said:
ok i tried it with that formula this way:

5 eV = p^2 / (9.11 x 10^-31kg)
p= 6.75 x 10^-17

this gave me a wavelength of 9.82 x 10^-18 m and wasn't correct...I'm still doing something wrong

You probably didn't converted energy from eV in Joules. And the formula is:

E=\frac{p^2}{2m_{e}}

(2 is missing in yours). I got the result \lambda=5.48\cdot{10^{-10}}m.
 
Last edited:
the 2 and converting to joules were the problems.

Much appreciated, I've got it now.
 
what would i do for the problem with E=415V since I can't just convert that into Joules?
 
The potential difference is 415v. This is not an energy. Recall the definition of the electron-volt...
 
It would be very useful if you studied theory once again before trying to solve any more problems, becoase if you have problems on this level it means that you haven't quite got the grasp of meaning of things and you will not be able to make any progress.
 
  • #10
thank you for the tips, I figured it out on reviewing the theory. Misunderstood the 415V, its been awhile since I worked with them:rolleyes:
 

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