Electrons with de-Broglie wavelength $\lambda$ fall on the target of an $X$-ray tube. The cut-off wavelength of the emitted $X$-ray is:

  • A
    $\lambda_0 = \frac{2mc\lambda^2}{h}$
  • B
    $\lambda_0 = \frac{2h}{mc}$
  • C
    $\lambda_0 = \frac{2m^2c^2\lambda^3}{h^2}$
  • D
    $\lambda_0 = \lambda$

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