Q 12-11-126JEE MainJEE Main 2021 (20 Jul, Shift 2)Easy
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:
Answer: (C) $\frac{2mc\lambda^2}{h}$
Electron kinetic energy: $K = \dfrac{p^2}{2m} = \dfrac{h^2}{2m\lambda^2}$. The shortest X-ray wavelength takes all of it:
$$\lambda_0 = \frac{hc}{K} = \frac{2mc\lambda^2}{h}$$
Solution by Sreeraj P, M.Sc Physics