Q. An electron of $K_{\alpha}$ spectral line of an atom is $59\ keV,$ then the wavelength of $K_{\alpha}$ line will be
A) $0.20\ \mathring{\mathrm{A}}$
B) $0.42\ \mathring{\mathrm{A}}$
C) $0.31\ \mathring{\mathrm{A}}$
D) $0.62\ \mathring{\mathrm{A}}$
Explanation
(a) $\lambda = \frac{hc}{eV} = \frac{12375}{59000} = 0.20\mathring{\mathrm{A}}\ \ \ \left\lbrack \because\frac{hc}{e} = 12375 \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.