Electrons of mass m with de-Broglie wavelengt | vedclass

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(C) The kinetic energy $K$ of an electron with de-Broglie wavelength $\lambda$ is given by $K = \frac{p^2}{2m}$.Since $p = \frac{h}{\lambda}$,we have

Vedclass · Accepted Answer

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

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(C) The kinetic energy $K$ of an electron with de-Broglie wavelength $\lambda$ is given by $K = \frac{p^2}{2m}$.Since $p = \frac{h}{\lambda}$,we have $K = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$.In an $X$-ray tube,the maximum energy of an emitted $X$-ray photon is equal to the kinetic energy of the incident electron.Therefore,$\frac{hc}{\lambda_0} = K = \frac{h^2}{2m\lambda^2}$.Solving for $\lambda_0$,we get $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an $X$-ray tube. The cutoff wavelength $(\lambda_0)$ of the emitted $X$-ray is

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