Bragg's law for $X$-rays is

The energy of an $X$-ray photon of wavelength $1.65 \mathring{A}$ is ....... $keV$. $(h = 6.6 \times 10^{-34} \ J \cdot s, \ c = 3 \times 10^8 \ m/s, \ 1 \ eV = 1.6 \times 10^{-19} \ J)$ (in $.5$)

The potential difference applied to an $X$-ray tube is $5 \text{ kV}$ and the current through it is $3.2 \text{ mA}$. Then the number of electrons striking the target per second is

If the energy of a $K$-shell electron is $-40000 \ eV$ and a potential of $60000 \ V$ is applied to a Coolidge tube,which of the following $X$-rays will be formed?

Frequency of the $K_{\alpha}$ $X$-ray for the element calcium $(Ca, Z = 20)$ is $8.95 \times 10^{17} \ Hz$. The frequency of the $K_{\alpha}$ $X$-ray for the element cadmium $(Cd, Z = 48)$ will be:

If the potential difference applied across an $X$-ray tube is $V$ volts,then the approximate minimum wavelength of the emitted $X$-rays will be: