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

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SUMMARY

The discussion focuses on calculating the velocity of an electron that possesses kinetic energy equivalent to the energy of a photon from sodium light at a wavelength of 590 nm. The photon energy is calculated using the equation E = hf, resulting in E = 3.371186441 × 10^-19 Joules. The kinetic energy equation E_{kinetic} = (mv^2)/2 is utilized to find the velocity, leading to the formula v = sqrt(2E/m). The participants clarify the correct application of units and constants, emphasizing the importance of using the electron mass in kilograms for accurate calculations.

PREREQUISITES
  • Understanding of the photoelectric effect and photon energy calculations
  • Familiarity with kinetic energy equations in classical mechanics
  • Knowledge of unit conversions between Joules and electronvolts
  • Basic grasp of wave-particle duality in quantum mechanics
NEXT STEPS
  • Study the derivation of the photon energy equation E = hf
  • Learn about the relationship between wavelength and frequency in electromagnetic waves
  • Explore the implications of mass-energy equivalence in particle physics
  • Investigate the concept of wave-particle duality and its applications in quantum mechanics
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Students in physics, particularly those studying quantum mechanics and electromagnetism, as well as educators looking for practical examples of photon energy and kinetic energy relationships.

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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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