Compare the wavelengths of a photon and an electron

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The discussion compares the wavelengths of a 2.2 eV photon and a 2.2 eV electron, revealing that the electron has a shorter wavelength (8.27 x 10^-10 m) compared to the photon (5.67 x 10^-7 m). This is attributed to the differences in momentum, as the electron's rest mass contributes to a greater momentum at the same energy level, leading to a shorter wavelength. The calculations for both particles are confirmed as correct, emphasizing that the speed of light does not directly correlate with wavelength in this context. The relationship between energy, momentum, and wavelength is clarified, noting that the electron's momentum increases more rapidly with energy than that of the photon. Understanding these principles is essential for applications like electron microscopy, which leverages the shorter wavelengths of electrons for higher resolution.
chef99
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Homework Statement


Compare and contrast a 2.2 eV photon with a 2.2 eV electron in terms of wavelength (m).[/B]

Homework Equations



p = h/λ

λ = h/mv

The Attempt at a Solution



For photon:

p = h/λ

λ = h/p

λ = (6.63 x10-34) / (1.17 x10-27kgm/s)**

λ = 5.67 x10-7 m

**I have already determined the momentum as is just one part of the question, in which the energy, momentum, rest mass and velocities are calculated.

For electron:

λ = h/mv

λ = (6.63 x10-34) / (9.11 x10-31kg)(8.8 x105m/s)

λ = 8.27 x10-10 m
I am confident in my answers, I am just curious as to why the wavelength for the electron is smaller than that of the photon, as I would have assumed the wavelength of the photon would be smaller, due to its greater velocity. I thought smaller wavelength means a higher frequency, which, in turn, means the wave is moving faster? However, this cannot be the case as the photon is obviously moving faster. I'd appreciate any explanation as I haven't been able to find one. This is, of course, assuming I've done my calculations correctly but I think I have. Thanks
 
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chef99 said:

Homework Statement


Compare and contrast a 2.2 eV photon with a 2.2 eV electron in terms of wavelength (m).[/B]

Homework Equations



p = h/λ

λ = h/mv

The Attempt at a Solution



For photon:

p = h/λ

λ = h/p

λ = (6.63 x10-34) / (1.17 x10-27kgm/s)**

λ = 5.67 x10-7 m

**I have already determined the momentum as is just one part of the question, in which the energy, momentum, rest mass and velocities are calculated.

For electron:

λ = h/mv

λ = (6.63 x10-34) / (9.11 x10-31kg)(8.8 x105m/s)

λ = 8.27 x10-10 m
I am confident in my answers, I am just curious as to why the wavelength for the electron is smaller than that of the photon, as I would have assumed the wavelength of the photon would be smaller, due to its greater velocity. I thought smaller wavelength means a higher frequency, which, in turn, means the wave is moving faster? However, this cannot be the case as the photon is obviously moving faster. I'd appreciate any explanation as I haven't been able to find one. This is, of course, assuming I've done my calculations correctly but I think I have. Thanks
Your calculations are correct. When speaking of the energy of the electron, it usually means the kinetic energy, From an electron and a photon of the same energy, the electron has greater momentum, and shorter wavelength. That is the basics of the electron microscopes
"An electron microscope is a microscope that uses a beam of accelerated electrons as a source of illumination. As the wavelength of an electron can be up to 100,000 times shorter than that of visible light photons, electron microscopes have a higher resolving power than light microscopes and can reveal the structure of smaller objects.ˇ" (Wikipedia)
The wavelength is inversely proportional to the momentum.
The momentum p=mv both for the electron and the photon, if m is the relativistic mass. For kinetic energies as low as a few eV, the mass of the electron is 9.1x10-31kg.
Because of the mass-energy relationship E=mc2, you can attribute the mass E/c2 to the photon, and now it is much less than the mass of the electron. In spite of the greater velocity, the momentum of the photon is smaller than that of the electron.
 
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You seem to have a couple of common misconceptions about waves, so here's a few things to think about.
chef99 said:
I am confident in my answers, I am just curious as to why the wavelength for the electron is smaller than that of the photon, as I would have assumed the wavelength of the photon would be smaller, due to its greater velocity.
Consider two photons with different energies. They both move at the speed of light yet they have different wavelengths. You don't want to relate speed to wavelength.

I thought smaller wavelength means a higher frequency, which, in turn, means the wave is moving faster?
Again, consider two photons with different frequencies. Both move at the same speed.

Also think about a wave on a string. Its speed is given by ##v = \sqrt{T/\mu}## where ##T## is the tension in the string and ##\mu## is the mass per unit length. The speed doesn't depend on wavelength or frequency.

However, this cannot be the case as the photon is obviously moving faster. I'd appreciate any explanation as I haven't been able to find one. This is, of course, assuming I've done my calculations correctly but I think I have. Thanks
For both a photon and an electron, you have ##\lambda = h/p##, so the difference in the wavelengths is due to the fact they have different momenta. For a photon, the energy and momentum are related by
$$p_\gamma = \frac{E}{c} = \frac{\sqrt{E^2}}c.$$ For a non-relativistic electron, you have
$$p_e = \sqrt{2mE} = \frac{\sqrt{2(mc^2)E}}{c} = \frac{\sqrt{2E_0 E}}c,$$ where ##E_0## is the rest energy of the electron. Comparing the expressions for the momentum of the photon and the electron, you can see that the rest energy of the electron causes the momentum of the electron to grow more quickly with energy than the momentum of the photon does, resulting in a shorter wavelength for the same kinetic energy.
 
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Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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