Q. Electrons of mass m with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength ($\lambda_0$) of the emitted X-ray is :-
A) $\lambda_0 = \frac{2mc\lambda^2}{h}$
B) $\lambda_0 = \frac{2h}{mc}$
C) $\lambda_0 = \frac{2m^2c^2\lambda^3}{h^2}$
D) $\lambda_0 = \lambda$
Explanation
[Solution] Electron KE $= \frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_0}$, so $\lambda_0 = \frac{2mc\lambda^2}{h}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.