I am struggling with getting this question done and need some help solving it, explain and make sure the answer is 100% correct.

When a fast electron (i.e., one moving at a relativistic speed) passes by a heavy atom, it interacts with the atom's electric field. As a result, the electron's kinetic energy is reduced; the electron slows down. In the meantime, a photon of light is emitted. The kinetic energy lost by the electron equals the energy Eγ�� of a photon of radiated light:

Eγ=K−K′��=�−�′,

where K� and K′�′ are the kinetic energies of the electron before and after radiation, respectively.

This kind of radiation is called bremsstrahlung radiation, which in German means "braking radiation" or "deceleration radiation." The highest energy of a radiated photon corresponds to the moment when the electron is completely stopped.

Part A.

Given an electron beam whose electrons have kinetic energy of 4.00 keVkeV , what is the minimum wavelength λmin�min of light radiated by such beam directed head-on into a lead wall?

Express your answer numerically in nanometers.

Part B .

How fast would an electron have to move so that its de Broglie wavelength would be 2.50 mmmm ?

Express your answer in meters per second.

Transcribed Image Text:How fast would an electron have to move so that its de Broglie wavelength would be 2.50 mm ? Express your answer in meters per second. || ΑΣΦ v= 2.90.107 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining m/s
Transcribed Image Text:When a fast electron (i.e., one moving at a relativistic speed) passes by a heavy atom, it interacts with the atom's electric field. As a result, the electron's kinetic energy is reduced; the electron slows down. In the meantime, a photon of light is emitted. The kinetic energy lost by the electron equals the energy Ey of a photon of radiated light: E = K - K', where K and K' are the kinetic energies of the electron before and after radiation, respectively. This kind of radiation is called bremsstrahlung radiation, which in German means "braking radiation" or "deceleration radiation." The highest energy of a radiated photon corresponds to the moment when the electron is completely stopped. Part A Given an electron beam whose electrons have kinetic energy of 4.00 keV, what is the minimum wavelength Amin of light radiated by such beam directed head-on into a lead wall? Express your answer numerically in nanometers. ► View Available Hint(s) Amin Submit V 0.206 ΑΣΦ Previous Answers X Incorrect; Try Again; 4 attempts remaining ? nm

Question

I am struggling with getting this question done and need some help solving it, explain and make sure the answer is 100% correct.

When a fast electron (i.e., one moving at a relativistic speed) passes by a heavy atom, it interacts with the atom's electric field. As a result, the electron's kinetic energy is reduced; the electron slows down. In the meantime, a photon of light is emitted. The kinetic energy lost by the electron equals the energy Eγ�� of a photon of radiated light:

Eγ=K−K′��=�−�′,

where K� and K′�′ are the kinetic energies of the electron before and after radiation, respectively.

This kind of radiation is called bremsstrahlung radiation, which in German means "braking radiation" or "deceleration radiation." The highest energy of a radiated photon corresponds to the moment when the electron is completely stopped.

Part A.

Given an electron beam whose electrons have kinetic energy of 4.00 keVkeV , what is the minimum wavelength λmin�min of light radiated by such beam directed head-on into a lead wall?

Express your answer numerically in nanometers.

Part B .

How fast would an electron have to move so that its de Broglie wavelength would be 2.50 mmmm ?

Express your answer in meters per second.

When a fast electron (i.e., one moving at a relativistic speed)
passes by a heavy atom, it interacts with the atom's electric field.
As a result, the electron's kinetic energy is reduced; the electron
slows down. In the meantime, a photon of light is emitted. The
kinetic energy lost by the electron equals the energy Ey of a
photon of radiated light:
E = K - K',
where K and K' are the kinetic energies of the electron before
and after radiation, respectively.
This kind of radiation is called bremsstrahlung radiation, which in
German means "braking radiation" or "deceleration radiation."
The highest energy of a radiated photon corresponds to the
moment when the electron is completely stopped.
Part A
Given an electron beam whose electrons have kinetic energy of 4.00 keV, what is the minimum wavelength Amin of light radiated by such beam directed
head-on into a lead wall?
Express your answer numerically in nanometers.
► View Available Hint(s)
Amin
Submit
V
0.206
ΑΣΦ
Previous Answers
X Incorrect; Try Again; 4 attempts remaining
?
nm

Accepted Answer

GivenWhen a fast electron (i.e., one moving at a relativistic speed) passes by a heavy atom, it…