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 :
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- 0
- \(\dfrac{2m^2c^2\lambda^2}{h^2}\)
- \(\dfrac{2mc\lambda^2}{h}\)
- \(\dfrac{hc}{mc}\)
The Correct Option is C
Solution and Explanation
Kinetic energy of the electron: \(K = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}\).
Minimum (cut-off) wavelength occurs when the electron gives all its energy to a single photon: \(K = \dfrac{hc}{\lambda_{min}}\).
\(\lambda_{min} = \dfrac{hc}{K} = \dfrac{hc\cdot 2m\lambda^2}{h^2} = \dfrac{2mc\lambda^2}{h}\).
Answer: (3) \(\dfrac{2mc\lambda^2}{h}\)
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