Question 387: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

Dual Nature of Matter and Radiation

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

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