During an x-ray exposure, an incident electron strikes the anode with 110 keV and passes very close to the tungsten nucleus. What is the approximate energy of the photon created by this interaction?

During an x-ray exposure, an incident electron strikes the anode with 110 keV and passes very close to the tungsten nucleus. What is the approximate energy of the photon created by this interaction?

• A. 0 keV
• B. 69 keV
• C. >110 keV
• D. 110 keV

Answer: (D)

Bremsstrahlung photons come from a fast electron losing energy to the strong electric field of a nucleus. The energy of the emitted photon comes from the electron’s kinetic energy, and it cannot exceed that energy. When the electron passes very close to the tungsten nucleus, its deceleration is maximal, so it can transfer nearly all of its 110 keV into radiation. Therefore the photon energy is approximately 110 keV. In reality the spectrum is continuous up to that maximum, but the scenario described points to the highest possible energy transfer, making 110 keV the best match.

Bremsstrahlung photons come from a fast electron losing energy to the strong electric field of a nucleus. The energy of the emitted photon comes from the electron’s kinetic energy, and it cannot exceed that energy. When the electron passes very close to the tungsten nucleus, its deceleration is maximal, so it can transfer nearly all of its 110 keV into radiation. Therefore the photon energy is approximately 110 keV. In reality the spectrum is continuous up to that maximum, but the scenario described points to the highest possible energy transfer, making 110 keV the best match.