In $X$-ray spectrum,the wavelength $\lambda$ of the $K_{\alpha}$ line depends on the atomic number $Z$ as:

$\Delta \lambda$ is the difference between the wavelength of the $K_\alpha$ line and the minimum wavelength of the continuous $X$-ray spectrum when the $X$-ray tube is operated at a voltage $V$. If the operating voltage is changed to $V / 3$,then the above difference is $\Delta \lambda^{\prime}$. Then:

When the accelerating voltage applied to the electrons is increased beyond a critical value in an $X$-ray tube,what happens to the $X$-ray spectrum?

$A$ potential difference of $20 \, kV$ is applied across an $X-ray$ tube. The minimum wavelength of $X-rays$ generated (in $\mathring{A}$) is

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:

The energy of a photon of light with wavelength $5000 \, \mathring{A}$ is approximately $2.5 \, eV$. In this way,the energy of an $X$-ray photon with wavelength $1 \, \mathring{A}$ would be: