Electrons of mass m with de-Broglie wavelength fall on the target in an X-ray tube. The…

Physics · NEET · NTA ExamsDual Nature of Matter and Radiation

Electrons of mass m with de-Broglie wavelength \(\lambda\) fall on the target in an X-ray tube. The cutoff wavelength \(\left(\lambda_{g}\right)\) of the emitted X-ray is:
  1. \(\lambda_{a}=\frac{2 m^{2} c^{2} \lambda^{3}}{h^{2}}\)
  2. \(\lambda_{g}=\lambda\)
  3. \(\lambda_{a}=\frac{2 m c \lambda^{2}}{h}\)
  4. \(\lambda_{a}=\frac{2 h}{m c}\)

Answer

(A) _ a = 2 m^ 2 c^ 2 ^ 3 h^ 2

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