After passing through a gold film of thickness $5 \times 10^{-3} \ m$,the intensity of an $X$-ray beam decreases to $36.8\%$ of its initial intensity. The absorption coefficient is ....... $m^{-1}$.

The primary difference between $X$-rays and $\gamma$-rays is that:

An electron having de-Broglie wavelength $\lambda$ is incident on a target in an $X$-ray tube. The cut-off wavelength of the emitted $X$-ray is:

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 $keV$?

If the wavelength of the $K_\alpha$ line of an element with atomic number $43$ is $\lambda$,then the wavelength of the $K_\alpha$ line of an element with atomic number $29$ is:

An $X$-ray tube with a copper target emits a $Cu$ $K_{\alpha}$ line of wavelength $1.50 \ \mathring{A}$. What should be the minimum voltage through which electrons are to be accelerated to produce this wavelength of $X$-rays? $(h = 6.63 \times 10^{-34} \ \text{J}\cdot\text{s}, c = 3 \times 10^8 \ \text{m/s})$