X-ray cut-off wavelength and de Broglie wavelength

2016 Phase II 1 question

1. 1 Dual Nature of Radiation and Matter
   Electrons of mass m with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength ($\lambda_0$) of the emitted X-ray is:
   A $\lambda_0 = \frac{2m^2c^2\lambda^3}{h^2}$ B $\lambda_0 = \lambda$ C $\lambda_0 = \frac{2mc\lambda^2}{h}$ D $\lambda_0 = \frac{2h}{mc}$
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