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Electrons with de-Broglie wave length $$\lambda$$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is
$$\lambda_0 = \frac{2mc \lambda^2}{h}$$
$$\lambda_0 = \frac{2h}{mc}$$
$$\lambda_0 = \frac{2 m^2 c^2 \lambda^3}{h^2}$$
$$\lambda_0 = \lambda$$
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