$A$ potential difference of $42,000 \ V$ is used in an $X-ray$ tube to accelerate electrons. The maximum frequency of the $X-radiations$ produced is ($1 \ eV = 1.6 \times 10^{-19} \ J$ and $h = 6.63 \times 10^{-34} \ J \cdot s$).

The maximum distance between the inter-atomic planes of a crystal is $10^{-7} \ cm$. The maximum wavelength of $X$-rays that can be studied by this crystal is ........... $\mathring{A}$.

Bragg's law for $X$-rays is

The $K_{\alpha}$ characteristic $X$-ray refers to the transition:

The wavelength of $L_{\alpha}$ line in $X$-ray spectrum of $_{78}Pt$ is $1.32 \ \mathring{A}$. The wavelength of $L_{\alpha}$ line in $X$-ray spectrum of another unknown element is $4.17 \ \mathring{A}$. If the screening constant for the $L_{\alpha}$ line is $7.4$,then the atomic number of the unknown element is:

If the frequency of the $K_\alpha$ line of an element with $Z = 64$ is $\nu_\alpha$ and the frequency of the $K_\alpha$ line of an element with $Z = 80$ is $\nu'_\alpha$,then the ratio of their frequencies is: