An X-ray tube with a target of atomic number $Z$ is operated at an accelerating voltage $V$. Let $\l

Question

An X-ray tube with a target of atomic number $Z$ is operated at an accelerating voltage $V$. Let $\lambda_c$ be the short-wave cutoff wavelength of the continuous spectrum, and $\lambda_K$ be the wavelength of the $K_{\alpha}$ characteristic line. Which of the following is/are correct?

Professor PC Thomas Classes · Answer

•	(A) Cutoff wavelength $\lambda_c = hc / (eV) \propto 1/V$. Doubling $V$ halves $\lambda_c$. Characteristic lines depend only on the target material ($Z$), so $\lambda_K$ is constant.
•	(B) $\Delta \lambda = \lambda_K - \lambda_c$. As $V$ increases, $\lambda_c$ decreases, so $\Delta \lambda$ increases, but the relationship is $A - B/V$, which is NOT linear with respect to $V$.
•	(C) $K_{\alpha}$ energy of Zn is $13.6 	imes (30-1)^2 	imes (3/4)$. The binding energy of the K-shell of Fe is roughly $13.6 	imes (26-1)^2$. Since $Z_{Zn} > Z_{Fe}$, the $K_{\alpha}$ photon from Zn is more energetic than the K-shell binding energy of Fe, so it can cause photoelectric emission.
•	(D) Moseley's law specifies $\omega_{K_{\alpha}} = \frac{3}{4} R(Z-\sigma)^2$ where $\sigma \approx 1$ for K-lines.