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(C) Let $K$ be the kinetic energy of the electron.The de-Broglie wavelength of the electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$

Vedclass · Accepted Answer

$\lambda_0 = \frac{2mc\lambda^2}{h}$

Vedclass · Answer

(C) Let $K$ be the kinetic energy of the electron.The de-Broglie wavelength of the electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mK}$,which implies $K = \frac{h^2}{2m\lambda^2}$.The cut-off wavelength $\lambda_0$ for continuous $X$-rays is determined by the maximum energy of the photon,which equals the kinetic energy of the incident electron: $E = \frac{hc}{\lambda_0} = K$.Substituting the value of $K$,we get $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.Solving for $\lambda_0$,we find $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.Therefore,option $(C)$ is correct.

An electron having de-Broglie wavelength $\lambda$ is incident on a target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-ray is:

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