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Physics · PYQ NEET UG 2016(2)
हिन्दीElectrons of mass \( m \) with de-Broglie wavelength \( \lambda \) fall on the target in an X-ray tube. The cut-off wavelength (\( \lambda_0 \)) of the emitted X-ray is
Correct answer · Option 2
\( \lambda_0 = \frac{2mc\lambda^2}{h} \)
Explanation
Kinetic energy \( K = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2} \). The maximum energy of X-ray photon is \( E = \frac{hc}{\lambda_0} = K \). Thus, \( \lambda_0 = \frac{2mc\lambda^2}{h} \).
