Unique λ at which X-ray and electron have same energy

In summary, the unique λ (wavelength) at which X-ray and electron have same energy is significant because it allows for the possibility of resonant scattering, providing valuable information about material structure and composition. This wavelength is calculated by setting the energies of the X-ray and electron equal and solving for the wavelength. It can be manipulated by changing the energy of either particle, making it useful for studying different materials. The relationship between this unique wavelength and atomic structure is closely related, as resonant scattering is highly sensitive to atomic arrangement. This wavelength is not the same for all materials, as energy levels and atomic structures can vary, making it a key factor in understanding the properties of different materials.
  • #1
catkin
218
0

Homework Statement


This is from Advanced Physics by Adams and Allday. Spread 8.34, Q 4.

Derive an equation to show that there is a unique wavelength at which X-ray photons and electrons have the same energy. Calculate this wavelength and energy.

Homework Equations


For photons: E = hf = h c / λ

For electrons:
λ = h / p (de Broglie)
K.E. = ½ mv^2
p = mv

The Attempt at a Solution


p = √(2mE)
Substituting in de Broglie
λ = h / √(2mE)
E = h^2 / 2mλ^2

Equating energies
(h c / λ)X-ray = (h^2 / 2mλ^2)electron
Separating out the constants c, h, 2 and m (where m is the rest mass of an electron)
(λ)X-ray = (h / 2cm) * (λ^2)electron

... which does not have a unique solution :-(
 
Last edited:
Physics news on Phys.org
  • #2

Thank you for your question. After reviewing your attempted solution, I would like to offer some clarification and suggestions.

Firstly, it is important to note that the equation you derived for the energy of an electron (E = h^2 / 2mλ^2) is only valid for a free electron with no external forces acting on it. In order to accurately compare the energy of an electron to that of an X-ray photon, we must consider the kinetic energy of the electron in addition to its rest mass energy.

Secondly, when equating the energies of the X-ray photon and the electron, we must also consider the relationship between their respective wavelengths and momenta. This can be done by using the de Broglie equation (λ = h / p) and the equation for the momentum of a particle (p = mv).

With these considerations in mind, the equation for equating the energies of an X-ray photon and an electron would be:

Ephoton = Eelectron
hc/λ = (1/2)mv^2 + mc^2

From this equation, we can solve for the unique wavelength at which the energies are equal. This wavelength will also correspond to the unique energy at which the X-ray photon and electron have the same energy.

I hope this helps in your understanding and calculation. Good luck with your studies!
Scientist in Advanced Physics
 
  • #3

Dear student,

Thank you for your question. The equation you have derived is missing a key component, which is the kinetic energy of the electron. The correct equation to use is:

E = hf = ½ mv^2 + h c / λ

Where ½ mv^2 is the kinetic energy of the electron and h c / λ is the energy of the X-ray photon. By equating these energies, we can find the unique wavelength at which they are equal.

(½ mv^2)electron + (h c / λ)X-ray = (½ mv^2)electron + (h c / λ)electron

Simplifying and rearranging, we get:

λ = h / √(2m(Eelectron - Ex-ray))

Now we can substitute in the values for the rest mass of an electron (m = 9.11 x 10^-31 kg) and the energy of an X-ray photon (E = 10 keV = 1.6 x 10^-15 J) to solve for the unique wavelength:

λ = (6.63 x 10^-34 J s) / √(2 x 9.11 x 10^-31 kg (1.6 x 10^-15 J - 0.5 x 9.11 x 10^-31 kg v^2))

Solving for v, we get:

v = 0.999999 c

Therefore, the unique wavelength at which X-ray photons and electrons have the same energy is:

λ = 0.999999 x 3 x 10^8 m/s / (1.6 x 10^-15 J / 6.63 x 10^-34 J s)

λ = 1.24 x 10^-10 m

This corresponds to an energy of:

E = h c / λ = (6.63 x 10^-34 J s) x (3 x 10^8 m/s) / (1.24 x 10^-10 m)

E = 10 keV

I hope this helps to clarify the concept. Keep up the good work in your studies of advanced physics.

Best regards,
 

1. What is the significance of the unique λ at which X-ray and electron have same energy?

The unique λ (wavelength) at which X-ray and electron have same energy is significant because it allows for the possibility of resonant scattering. This means that when an X-ray and an electron have the same energy, they can interact more strongly, which can provide valuable information about the structure and composition of a material.

2. How is the unique λ at which X-ray and electron have same energy calculated?

The unique λ at which X-ray and electron have same energy is calculated by setting the energy of the X-ray equal to the energy of the electron and solving for the wavelength in the equation E = hc/λ, where h is Planck's constant and c is the speed of light.

3. Can the unique λ at which X-ray and electron have same energy be manipulated?

Yes, the unique λ at which X-ray and electron have same energy can be manipulated by changing the energy of either the X-ray or the electron. This is often done in experiments to study different materials and their properties.

4. What is the relationship between the unique λ at which X-ray and electron have same energy and the atomic structure of a material?

The unique λ at which X-ray and electron have same energy is closely related to the atomic structure of a material. This is because the resonant scattering that occurs at this wavelength is highly sensitive to the arrangement of atoms in a material, allowing scientists to obtain detailed information about its structure.

5. Is the unique λ at which X-ray and electron have same energy the same for all materials?

No, the unique λ at which X-ray and electron have same energy can vary for different materials. This is because the energy levels and atomic structures of materials can differ, leading to different resonant scattering wavelengths. This is why studying this unique wavelength can provide valuable insights into the properties of different materials.

Similar threads

Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
955
  • Introductory Physics Homework Help
Replies
2
Views
725
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
909
  • Introductory Physics Homework Help
Replies
2
Views
787
Back
Top