Finding KE when given de Broglie wavelength

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SUMMARY

The discussion focuses on calculating the kinetic energy (KE) of a proton given its de Broglie wavelength of 10 femtometers (fm). The correct formula used is the de Broglie wavelength equation, λ = h/(mv), and the kinetic energy formula, KE = (1/2)(m)(v^2). The participants identified a common error in unit conversion, emphasizing the necessity of maintaining SI units throughout the calculations to achieve accurate results. The final kinetic energy was calculated to be approximately 8.2 x 10^-18 MeV.

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  • Understanding of de Broglie wavelength and its significance in quantum mechanics
  • Familiarity with kinetic energy calculations in classical mechanics
  • Knowledge of unit conversions between MeV, eV, and Joules
  • Proficiency in using fundamental constants such as Planck's constant (h)
NEXT STEPS
  • Study the derivation and applications of the de Broglie wavelength equation
  • Learn about the relationship between mass, velocity, and kinetic energy in classical physics
  • Explore unit conversion techniques between different energy units like MeV, eV, and Joules
  • Investigate the implications of quantum mechanics on particle behavior at small scales
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Students in physics, particularly those studying quantum mechanics and classical mechanics, as well as educators looking to clarify concepts related to kinetic energy and wave-particle duality.

vspectra
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[SOLVED] Finding KE when given de Broglie wavelength

Homework Statement


What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of 10fm?
1 MeV = 10^6 eV
1 fm = 10^-15m
1 eV = 1.602 x 10^-19 J
h = 6.63 x 10^-34 Js
wavelength = 10 x 10^-15 m = 10^-5 nm
Mass proton = 1.673 x 10^-27 kg


Homework Equations



wavelength = h/mv
KE = (1/2)(m)(v^2)


The Attempt at a Solution



I feel like I did this right, but I still got the incorrect answer and I can't seem to notice what I did wrong.

v = (6.63 x 10^-34) / (10^-5)(1.673 x 10^-27) = .03962 m/s
K = .5(1.673 x 10^-27)(.03862^2) = 1.316 x 10^-30 J = 8.2 x 10^-12 eV
= 8.2 x 10^-18 MeV
 
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Welcome to PF,
vspectra said:
v = (6.63 x 10^-34) / (10^-5)(1.673 x 10^-27) = .03962 m/s
K = .5(1.673 x 10^-27)(.03862^2) = 1.316 x 10^-30 J = 8.2 x 10^-12 eV
= 8.2 x 10^-18 MeV
Shouldn't that be 10*10-15? :wink:
 
Hootenanny said:
Welcome to PF,

Thanks.

Hootenanny said:
Shouldn't that be 10*10-15? :wink:



I tried that the first time I did the problem, and realized 10 x 10^-15 is in meters. Wouldn't I need the wavelength to be in nm? So it would be 1 x 10^-5 nm.
 
Last edited:
vspectra said:
I tried that the first time I did the problem, and realized 10 x 10^-15 is in meters. Wouldn't I need the wavelength to be in nm? So it would be 1 x 10^-5 nm.
Why would you want it in nm? If you want your output velocity to be in m/s, then all your inputs must be in SI units.
 
Ah, you're right, silly me. I was looking at another example which was basically the same question, and to me it looked like they converted from angstroms to nanometers for wavelength. It was actually converted to meters.

Thanks!
 


Hey thanks both of u :) i made the same mistake :p
 

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