Electrons with de-Broglie wavelength fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is [2007]

Question

Electrons with de-Broglie wavelength $λ$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is [2007]

Smart Achievers · Accepted Answer

(1)

The cut-off wavelength of the emitted X-ray,

$λ_{0} = \frac{h c}{e V}$ $\dots (i)$

According to de Broglie equation,

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}}$

or $λ^{2} = \frac{h^{2}}{2 m e V}$ $\Rightarrow V = \frac{h^{2}}{2 m e λ^{2}}$ $\dots (i i)$

From eq. (i) and (ii),

$λ_{0} = \frac{h c \times 2 m e λ^{2}}{e h^{2}} = \frac{2 m c λ^{2}}{h}$

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