De Broglie Wavelength for 1Mev electron

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SUMMARY

The de Broglie wavelength of a 1 MeV electron is calculated using the formula λ = h/p, where p is derived from the equation p = (2mT + T^2/c^2)^{1/2}. The correct calculation yields a wavelength of approximately 872 femtometers, contrary to the initial incorrect result of 4.09 x 10^{-21} femtometers. The error arose from using mass in energy units (eV) instead of the appropriate mass units.

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  • Knowledge of energy-mass equivalence
  • Basic proficiency in unit conversions, particularly between eV and kg
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Homework Statement



Calculate the de Broglie Wavelength of a 1 MeV electron. Express your answer in femptometers.

Homework Equations



## λ = h/p ##
## p = (2mT + T^2/c^2)^{1/2} ##

The Attempt at a Solution



basically just plugged everything in.

## λ = \frac{h}{(2mT + T^2/c^2)^{1/2}} ##

## λ = \frac{(4.126*10^{-15}eV*s)}{(2*(0.511*10^6 eV)(10^6 eV) + (10^6 eV)^2/c^2)^{1/2}} ##

## λ = 4.09*10^{-21} ##

Which is way too low. It's supposed to be ##872*10^{-15}##.
 
Last edited:
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Watch the units. You entered a mass in units of energy (eV).
 

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