Finding the deBroglie wavelength: conceptual issue

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SUMMARY

The discussion centers on calculating the de Broglie wavelength of an electron with a kinetic energy of 3.50 eV. The correct approach involves using the formula λ = h/(mv), where m is the mass of the electron and v is its velocity derived from kinetic energy. The user initially misapplied the total energy equation Etotal = K + Er, leading to an incorrect wavelength calculation of 0.00242 nm instead of the correct 0.656 nm. The confusion arose from incorrectly applying the relationship E = pc, which is valid only for massless particles like photons.

PREREQUISITES
  • Understanding of kinetic energy and its relation to velocity (K = ½mv²)
  • Familiarity with the de Broglie wavelength formula (λ = h/p)
  • Basic knowledge of relativistic energy concepts (E = mc²)
  • Concept of momentum in relation to mass and velocity (p = mv)
NEXT STEPS
  • Study the derivation and application of the de Broglie wavelength formula in quantum mechanics.
  • Learn about the differences between relativistic and non-relativistic equations for particles with mass.
  • Explore the implications of rest energy and kinetic energy in particle physics.
  • Investigate the conditions under which E = pc applies and its significance for massless particles.
USEFUL FOR

Students in physics, particularly those studying quantum mechanics and particle physics, as well as educators seeking to clarify concepts related to energy and wavelength calculations for particles with mass.

bachfromthedead
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Hello all! I am new to this forum, though I have been lurking for a long time. I intend to fill out my profile and introduce myself, but finals leave me with little time to spare at the moment. I have this final next week and would like to sort this out beforehand, hence the hasty post!
Thank you in advance.

1. Homework Statement

An electron has kinetic energy 3.50eV. Find it's wavelength.

Homework Equations


(1) p=mv
(2) K=½mv2
(3) p=E/c=h/λ

(4) Er=mc2
(5) Etotal=K+Er

The Attempt at a Solution


Note: my prof said not to use relativistic equations, as the energy is not high enough, but we have covered relativity in this class.
I now know how to solve this problem:
I solved for the speed of the electron with the kinetic energy formula, and put it into the following:
λ=h/(mv)
I got 0.656nm, which is correct.

My problem is that I did not initially solve the problem this way. What I did first was calculate the rest energy of the electron (Er=mc2) and add it to the kinetic energy to get the "total" energy. This was based off an equation which I was given by my prof: Etotal=K+Er. I then took that "total" energy and solved for the wavelength by deriving the equation: λ=hc/E from equation (3) above. I got the wrong answer, 0.00242nm, with this method.
What I don't understand is why my original method does not work. I'm assuming my fundamental understanding of these equations is wrong. Is the total energy of the electron not the rest energy plus the kinetic energy? Is there something I'm missing? My professor was not helpful this time around, so I'm turning to you guys!
Thank you for your time.
 
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Welcome to PF!

bachfromthedead said:
This was based off an equation which I was given by my prof: Etotal=K+Er. I then took that "total" energy and solved for the wavelength by deriving the equation: λ=hc/E from equation (3) above. I got the wrong answer, 0.00242nm, with this method.

In getting the equation λ=hc/E, it looks like you assumed that the momentum is related to energy by p = E/c. This is not correct for electrons. (But it is correct for photons.)
 
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[QUOTE="bachfromthedead, post: 5450436, member: 591656".[/QUOTE]
E = pc is true only if E is kinetic energy only. Rest energy must be zero, as in a photon.
For a particle of rest mass m, E = γmc2 - mc2 = kineic energy = total energy - rest energy.
 
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Thank you very much to both of you! It makes sense now, I think that the way these equations were presented in my class was the root of my confusion. I should have realized that E=pc would only apply to photons as it includes c.
 
bachfromthedead said:
I should have realized that E=pc would only apply to photons as it includes c.
For particles that have mass ##m##, ##E = \sqrt{(pc)^2 + (mc^2)^2}##. This still contains ##c##.

For massless particles like the photon, this reduces to ##E = pc##.
 

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