The $K_\alpha$ $X$-ray emission line for tungsten occurs at $\lambda = 0.021 \ nm$. What is the energy difference between the $K$ and $L$ levels in this atom in $keV$?

The wavelength of ${K_{\alpha }}$ $X$-rays produced by an $X$-ray tube is $0.76 \, \mathring{A}$. The atomic number of the anticathode material is:

If $X$-rays pass through a strong magnetic field,the $X$-rays . . . . . .

Calculate the wavelength of the $K_{\alpha}$ line for $Z=31$, when $a=5 \times 10^7 \, Hz^{1/2}$ for a characteristic $X$-ray spectrum.

$X$-rays incident on a material:

The $K_{\alpha}$ $X$-ray of molybdenum has a wavelength of $0.071 \, nm$. If the energy of a molybdenum atom with a $K$ electron knocked out is $27.5 \, keV$,the energy of this atom when an $L$ electron is knocked out will be $.... \, keV$. (Round off to the nearest integer) $[h = 4.14 \times 10^{-15} \, eVs, c = 3 \times 10^{8} \, ms^{-1}]$