Electrons with de-Broglie wavelength lambda f | vedclass

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Question

(A) The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.Squaring bot

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.Squaring both sides,we get $\lambda^2 = \frac{h^2}{2meV}$,which implies $eV = \frac{h^2}{2m\lambda^2}$.The cut-off wavelength $\lambda_0$ of the emitted $X$-rays corresponds to the maximum energy of the photon,which is equal to the kinetic energy of the incident electron: $E = eV = \frac{hc}{\lambda_0}$.Equating the two expressions for $eV$,we have $\frac{hc}{\lambda_0} = \frac{h^2}{2m\lambda^2}$.Solving for $\lambda_0$,we get $\lambda_0 = \frac{hc \cdot 2m\lambda^2}{h^2} = \frac{2mc\lambda^2}{h}$.

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

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