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

AI Thread Summary
To determine the speed an electron must move to match the kinetic energy of a photon from sodium light at 590 nm, the photon energy is calculated using E = hf, resulting in approximately 3.37 x 10^-19 joules. The kinetic energy equation, E_kinetic = mv^2/2, can be used to find the velocity by equating the photon energy to the kinetic energy. There is confusion regarding the correct formula and units, particularly with the mass of the electron and its representation in joules versus kilograms. After some calculations, a velocity of approximately 2.86 x 10^-3 m/s is suggested, but discrepancies in the equations and units are noted. Clarification on the proper use of units and equations is necessary for accurate results.
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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.
 
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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.
 
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Would it be like this

<br /> v = \frac{{\sqrt{{2E }} \div {m} }<br />
 
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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
 
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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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