Electrons with de-Broglie wavelength lambda | vedclass

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Question

for striking $\mathrm{e}^{-}: \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} 	herefore \mathrm{eV}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda^{2}}$

Vedclass · Accepted Answer

${\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}$

Vedclass · Answer

for striking $\mathrm{e}^{-}: \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} 	herefore \mathrm{eV}=\frac{\mathrm{h}^{2}}{2 \mathrm{m} \lambda^{2}}$

for $\mathrm{X}$ -ray $: \lambda_{\min }=\frac{\mathrm{hc}}{\mathrm{eV}}=\frac{\mathrm{hc}}{\mathrm{h}^{2} / 2 \mathrm{m} \lambda^{2}}=\frac{2 \mathrm{mc} \lambda^{2}}{\mathrm{h}}$

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

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