Electrons with de-Broglie wavelength lambda f | vedclass

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(A) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.Squaring both sides,we get

Vedclass · Accepted Answer

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

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(A) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $E$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$.Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mE}$.Rearranging for kinetic energy $E$,we find $E = \frac{h^2}{2m\lambda^2}$.The cut-off wavelength $\lambda_0$ of the emitted $X$-rays corresponds to the maximum energy of the photons,which is equal to the kinetic energy of the incident electrons: $E = \frac{hc}{\lambda_0}$.Substituting the expression for $E$,we have $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.Solving for $\lambda_0$,we get $\lambda_0 = \frac{2mc\lambda^2}{h}$.

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

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