JEE Advance - Physics (2007 - Paper 2 Offline - No. 9)

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

$$\lambda_{0}=\frac{2 m c \lambda^{2}}{h}$$

$$\lambda_{0}=\frac{2 h}{m c}$$

$$\lambda_{0}=\frac{2 m^{2} c^{2} \lambda^{3}}{h^{2}}$$

$$\lambda_{0}=\lambda$$

The cut off wavelength is given by

$$\lambda_{0}=\frac{h c}{e v}$$ ..... (i)

According to de-Broglie equation

$$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e \mathrm{~V}}}$$

$$\Rightarrow \quad \lambda^{2}=\frac{h^{2}}{2 m e \mathrm{~V}}$$

$$\Rightarrow \quad \mathrm{V}=\frac{h^{2}}{2 m e \lambda^{2}}$$ ..... (ii)

From (i) and (ii)

$$\lambda_{0}=\frac{h c \times 2 m e \lambda^{2}}{e h^{2}}=2 m c \lambda^{2}$$