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