Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target. The cut-off wavelength…
Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target. The cut-off wavelength $\lambda_0$ is equal to [ $h=$ Planck's constant, $C=$ velocity of light $]$
$\frac{2 m c \lambda^2}{h}$
$\frac{m c \lambda}{h}$
$\frac{2 h}{m c \lambda^2}$
$\frac{2 m c \lambda}{h}$
Solution
Using de-Broglie equation $\lambda=\frac{h}{p}$ where $p=\sqrt{2 m E}$
$\Rightarrow \lambda=\frac{h}{\sqrt{2 m E}}$
Energy of the X-ray emitted $E=\frac{h c}{\lambda_0}$
$\therefore \lambda=\frac{h}{\sqrt{2 m \times \frac{h c}{\lambda_0}}} \Rightarrow \lambda_0=\frac{2 m c \lambda^2}{h}$
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