What is the energy of a 500 nm photon and electron in quantum physics?

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SUMMARY

The energy of a 500 nm photon is calculated using the formula Eph = hf = hc/λ, resulting in an energy of 2.49 eV. For a non-relativistic electron with the same wavelength, the kinetic energy is determined using K = 1/2(mv²), yielding an energy of approximately 6.03E-6 eV. The momentum of the electron is calculated as p = h/λ, resulting in a value of 1.3E-27 J·m·s. The calculations demonstrate the distinct energy levels of photons and electrons at the same wavelength.

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Homework Statement


Calculate the energy (in eV) of a 500 nm photon. Calculate the non-relativistic kinetic energy for a 500 nm electron.

The Attempt at a Solution



Eph=hf=hc/λ = (6.63E-34 J·s)(3E8)/(500E-9 m) = 3.98E-19 J = 2.49 eV

I'm not sure why, but I thought that the energy of the electron would be the energy of the photon/c. However, that doesn't really make sense, nor does it yield the correct answer. The answer is 6.02E-6 eV. Any help on how I would find this would be appreciated.
Thank you.
 
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Can you find the momentum of a 500 nm electron?
 
p = h/λ
= (6.63E-34 J·s)/(500E-9m)
= 1.3eE-27 J·m·s

p = mv
v = 1455.5 m/s
K = 1/2(mv2)= 9.65E-25 J
= 6.03E-6 eV

Thank you. This should have been obvious.
 

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