Answer
See explanation.
Work Step by Step
a. Apply equation 41.48, relating the x-ray energy to Z. Here, Z=20.
$$E_{K_\alpha}=(20-1)^2(10.2\;eV)=3.7keV$$
The energy of a photon is related to its frequency.
$$f=\frac{E}{h}=\frac{(20-1)^2(10.2\;eV)}{4.136\times10^{-15}eV\cdot s}=8.9\times10^{17}Hz$$
Now find the wavelength.
$$\lambda=\frac{c}{f}=3.4\times10^{-10}m$$
b. Repeat for Z=27.
$$E_{K_\alpha}=(27-1)^2(10.2\;eV)=6.9keV$$
The energy of a photon is related to its frequency.
$$f=\frac{E}{h}=\frac{(27-1)^2(10.2\;eV)}{4.136\times10^{-15}eV\cdot s}=1.7\times10^{18}Hz$$
Now find the wavelength.
$$\lambda=\frac{c}{f}=1.8\times10^{-10}m$$
c. Repeat for Z=48.
$$E_{K_\alpha}=(48-1)^2(10.2\;eV)=22.5keV$$
The energy of a photon is related to its frequency.
$$f=\frac{E}{h}=\frac{(48-1)^2(10.2\;eV)}{4.136\times10^{-15}eV\cdot s}=5.4\times10^{18}Hz$$
Now find the wavelength.
$$\lambda=\frac{c}{f}=5.5\times10^{-11}m$$
