Answer
Copper.
Work Step by Step
Use the equation developed in Example 28-7 (page 818) to calculate the wavelength of the $K_{\alpha}$ line.
$$\frac{1}{\lambda}=(1.097\times10^7m^{-1})(Z-1)^2(\frac{1}{(n’)^2}-\frac{1}{n^2})$$
Solve for Z.
$$Z=1+\sqrt{\frac{1}{\lambda}(1.097\times10^7m^{-1})^{-1}(\frac{1}{(n’)^2}-\frac{1}{n^2})^{-1}}$$
The $K_{\alpha}$ line is for electrons moving from n = 2 down to n’ = 1.
$$Z=1+\sqrt{\frac{1}{0.154\times10^{-9}m}(1.097\times10^7m^{-1})^{-1}(\frac{1}{1^2}-\frac{1}{2^2})^{-1}}$$
$$Z=1+28.1\approx 29$$
The unknown material has Z=29, and corresponds to copper.