Answer
Chromium.
Work Step by Step
Use the equation developed in Example 28-7 (page 818) to calculate the wavelength of the $K_{\alpha}$ line.
$$\frac{1}{\lambda}=(1.097\times10^7m^{-1})(Z-1)^2(\frac{1}{(n’)^2}-\frac{1}{n^2})$$
The wavelength is inversely proportional to $(Z-1)^2$.
$$\frac{\lambda_{unknown}}{\lambda_{Fe}}=\frac{(Z_{Fe}-1)^2}{(Z_{unknown}-1)^2}$$
$$Z_{unknown}=((26-1)\sqrt{\frac{194pm}{229pm}})+1=24$$
The unknown material has Z=24, and corresponds to chromium.
