Question:
An electron having de-Broglie wavelength $\lambda$ is incident on a target in a X-ray tube. Cut-off wavelength of emitted X-ray is:
Correct Option: , 3
Solution:
$\lambda=\frac{h}{m V}$
kinetic energy, $\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{c}}}$
$\lambda_{\mathrm{C}}=\frac{2 \mathrm{~m} \lambda^{2} \mathrm{c}}{\mathrm{h}}$
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