Modern Physics: What is the wavelength of the most energetic photons produced by

Question

Modern Physics:

What is the wavelength of the most energetic photons produced by an X-ray tube using V0 = 219 kV potential difference?

What is the de Broglie wavelength of the electrons, just before they hit the anode inside the x-ray tube?

Both of these are two parts of the same question.

Main Menu Contents Grades Syllabus Course Contents » .. Homework for Topic 4. Wave Properties of Matter » High voltage X-ray tube (2) ·TimerNotes Evaluate Feedback-Print info What is the wavelength of the most energetic photons produced by an X-ray tube using Vo 219 kV potential difference? 567 pm Hints: How much energy do the electrons gain after they are accelerated through a potential difference Vo? Try to use electron-Volt units instead of Joules, hc = 1.2398 x106 eVpm. If the energy of one of those electrons is used up completely to create one single X-ra photon, then what is the frequency of that photon? If you know the frequency of the photon, how do you calculate its wavelength Submit Answer Incorrect. Tries 1/3 Previous Tries What is the de Broglie wavelength of the electrons, just before they hit the anode inside the x-ray tube? 1.89e-20 m Hints: What is the relationship between the de Broglie wavelength and the momentum of a particle? Can you express the momentum of the particle in terms of the kinetic energy and the rest energy of the particle? How did the electron acquire its kinetic energy? Submit Answer Incorrect. Tries 1/3 Previous Tries

Explanation / Answer

Given,

V = 219 kV

We know that

hc/lambda = eV

lambda = hc/eV

lambda = 6.6 x 10^-34 x 3 x 10^8/(1.6 x 10^-19 x 219 x 10^3) = 5.65 x 10^-12 m

Hence, lambda = 5.65 x 10^-12 m = 5.65 pico-m

KE = qV

KE = 1.6 x 10^-19 x 219 x 10^3 = 3.5 x 10^-14 J

pc = sqrt (2 KE m0c^2)

pc = sqrt (2 x 3.5 x 10^-14 x 9.1 x 10^-31 x (3 x 10^8)^2) = 7.57 x 10^-14 kg-m/s

lambda = hc/pc = 6.6 x 10^-34 x 3 x 10^8/(7.57 x 10^-14) = 2.62 x 10^-12 m

Hence,lambda = 2.62 x 10^-12 m = 2.62 pm

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