Here are the K\alpha wavelengths of a few elements:
\begin{array}{lccc} \hline \text { Element } & \lambda( pm ) & \text { Element } & \lambda( pm ) \\ \hline Ti & 275 & Co & 179 \\ V & 250 & Ni & 166 \\ Cr & 229 & Cu & 154 \\ Mn & 210 & Zn & 143 \\ Fe & 193 & Ga & 134 \\ \hline \end{array}
Make a Moseley plot (like that in Fig. 40-16) from these data and verify that its slope agrees with the value given for C in Module 40-6.
From the data given in the problem, we calculate frequencies (using Eq. 38-1), take their square roots, look up the atomic numbers (see Appendix F), and do a least-squares fit to find the slope: the result is 5.02 × 10^7 with the odd-sounding unit of a square root of a hertz. We remark that the least squares procedure also returns a value for the y-intercept of this statistically determined “best-fit” line; that result is negative and would appear on a graph like Fig. 40-17 to be at about – 0.06 on the vertical axis. Also, we can estimate the slope of the Moseley line shown in Fig. 40-17:
f=\frac{c}{\lambda} . (38-1)
\frac{(1.95-0.50) 10^9 Hz ^{1 / 2}}{40-11} \approx 5.0 \times 10^7 Hz ^{1 / 2}