Calculating Wavelength of Photons to Produce 30 keV Electrons

In summary, the question is asking for the initial wavelength of a photon in Compton scattering that will produce an electron with a kinetic energy of 30 keV. This can be calculated using the equation E=hc/λ, where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. The context of the discussion is important, as the phenomenon of Compton scattering and the de Broglie wavelength are relevant to finding the solution.
  • #1
KingNothing
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Homework Statement


56. What wavelength of photons are needed to produce 30 keV electrons in a compton scattering?


Homework Equations


[tex]E=\frac{hc}{\lambda}[/tex]

The Attempt at a Solution


This question throws me off because I was under the impression that compton scattering didn't produce any electrons, but simply deflected them. It also confuses me because I assume it's asking "What photons will produce electrons with total energy of 30-keV?" However, electrons have that much energy in rest mass alone.

Can anyone interpret this better?
 
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  • #2


Not sure what compton scattering is, but I'd find the velocity of the electrons from their energy and and then use the de Broglie wavelength formula


[tex]\lambda = \frac{h}{mv}[/tex]
 
  • #3


With all due respect, your suggestion does not make sense in the context of what we've discussed. We've discussed compton scattering, and have not discussed de Broglie wavelengths.
 
  • #4


Here is my interpretation:

Assume that you have an electron at rest. An incoming photon Compton scatters off it and the electron acquires a kinetic energy of 30 keV. What is the initial wavelength of the photon?
 
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Related to Calculating Wavelength of Photons to Produce 30 keV Electrons

1. How do I calculate the wavelength of photons to produce 30 keV electrons?

The formula for calculating the wavelength of photons is given by λ = hc/E, where λ is the wavelength in meters, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.00 x 10^8 m/s), and E is the energy of the photon in joules. To produce 30 keV electrons, we must convert 30 keV to joules by multiplying it by 1.6 x 10^-19 (the conversion factor). This gives us an energy of 4.8 x 10^-15 J. Plugging this value into the formula, we get a wavelength of approximately 2.48 x 10^-11 meters.

2. What is the relationship between energy and wavelength of photons?

The energy and wavelength of photons are inversely proportional. This means that as the energy of a photon increases, its wavelength decreases. This relationship is described by the formula E = hc/λ, where E is the energy in joules, h is Planck's constant, c is the speed of light, and λ is the wavelength in meters.

3. Can the wavelength of photons be adjusted to produce 30 keV electrons?

Yes, the wavelength of photons can be adjusted to produce 30 keV electrons. This can be done by changing the energy of the photons, either by adjusting the intensity of the light source or by using a different type of light source with a specific energy output.

4. What is the significance of producing 30 keV electrons?

Producing 30 keV electrons is significant because this energy level is often used in medical and industrial applications, such as X-ray imaging and radiation therapy. By accurately calculating the wavelength of photons needed to produce 30 keV electrons, we can ensure the effectiveness and safety of these applications.

5. Can other factors affect the accuracy of calculating the wavelength of photons?

Yes, there are several factors that can affect the accuracy of calculating the wavelength of photons. These include experimental error, uncertainties in the values of Planck's constant and the speed of light, and the effects of relativity at high energies. It is important to consider these factors and use the most precise and up-to-date values in order to obtain the most accurate calculation of the wavelength of photons.

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