Photons and matter waves: The photon of quantum light: I

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Homework Help Overview

The discussion revolves around calculating the velocity of an electron that has a kinetic energy equal to the energy of a photon from sodium light at a wavelength of 590 nm. The subject area includes concepts from quantum mechanics and classical mechanics, specifically relating to photon energy and kinetic energy equations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between photon energy and kinetic energy, considering the relevant equations. There are attempts to equate the two energies to solve for velocity. Some participants express confusion about the correct formulation of the equations and the units involved.

Discussion Status

Participants are actively discussing the calculations and equations needed to find the velocity. Some have provided guidance on how to set up the equations, while others are questioning the accuracy of the calculations and the proper use of units. There is no explicit consensus on the correct approach or final answer yet.

Contextual Notes

There are indications of potential confusion regarding the mass of the electron and its representation in different units. Participants are also addressing discrepancies in calculated values for photon energy, suggesting a need for careful verification of mathematical steps.

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Homework Statement


How fast must an electron move to have a Kinetic Energy equal to the photon energy of sodium light at wavelength 590 nm.


Homework Equations


photon energ E =hf

h = 6.63 * 10 ^ -34 J * s = 4.14 = 10^-15 eV *s

f = c/ lamda

c = 3 * 10^ 8 m/s

mass of an electron is 9.11 * 10 ^ -31 (kg) or 511 keV


The Attempt at a Solution



Ok I solved for the energy of a photon for sodium light

E = 6.63 * 10 ^ -34 J *s * ( (3* 10 ^ 8 m/s) / 590 * 10 ^ -9 meters) = 1.989 * 10 ^ -25 Joules




Were solving for Velocity. How do I do this problem it's confusing.
 
Last edited:
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Well you're looking for the kinetic energy (and thus velocity) right? Well E_{kinetic} = \frac{{mv^2 }}{2} and you know the energy of the photon so equate them and find the velocity.
 
Sorry about that I'll make my information much clearer next time. So that's the equation I use.
 
Last edited:
Would it be like this

<br /> v = \frac{{\sqrt{{2E }} \div {m} }<br />
 
Last edited:
Yes, where E = hf = hc/lamda
 
So is that E of photon energy of soidum light the same as the kinetic energy
 
Check your math again. I get a different value for energy using the same numbers you gave as input.
 
so then it would be


It would be v = sqrt { 2 (3.37 * 10 ^ -19 joules / 8.19 * 10^ -14 joules}

m of electron in joules is 8.19 * 10 ^ -14 joules
 
Last edited:
Oh yea the Enery of a photon for sodium light is actually
<br /> E = 3.371186441 \times 10^-19<br />
 
  • #10
That's what I got.
 
  • #11
I got 2.86 * 10 ^ -3 m/s
 
  • #12
That doesn't seem correct. Your equation for v doesn't look right, but I don't know if you are just typing in the tex code wrong. Did you really intend for the mass, m, to be in the numerator of that equation?
 
  • #13
Also, I'm not sure where you are getting the mass of the electron in joules. If you are using SI units of joules for energy, then the electron mass should be in kg.
 
  • #14
Wait yes, your equation for the velocity in terms of the energy changed... its divided by m.
 
  • #15
Well my last word on this...it looks like you threw another factor of c in there somewhere. If I divide the answer I get in meters/sec by 3*10^8 m/s, then I get the answer you posted above. So your answer would be correct if it was in units of "c".
 

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