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(D) The de-Broglie wavelength of the electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.The kinetic energy $K$ of

Vedclass · Accepted Answer

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

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(D) The de-Broglie wavelength of the electron is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.The kinetic energy $K$ of the electron is $K = \frac{p^2}{2m}$.Substituting $p = \frac{h}{\lambda}$,we get $K = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$.In an $X$-ray tube,the maximum energy of the emitted $X$-ray photon (corresponding to the cut-off wavelength $\lambda_c$) is equal to the kinetic energy of the incident electron: $K = \frac{hc}{\lambda_c}$.Equating the two expressions for $K$: $\frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_c}$.Solving for $\lambda_c$: $\lambda_c = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

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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