The binding energy of the innermost electron is $40 \, keV$. To obtain characteristic $X$-rays using a tungsten target in an $X$-ray tube,the potential difference $V$ between the cathode and the anode must be:

An electron beam is accelerated by a potential difference $V$ to hit a metallic target to produce $X$-rays. It produces continuous as well as characteristic $X$-rays. If $\lambda_{\min}$ is the smallest possible wavelength of $X$-ray in the spectrum,the variation of $\log \lambda_{\min}$ with $\log V$ is correctly represented in:

The wavelength of the $L_\alpha$ $X$-ray line of $Pt^{78}$ is $1.32 \, \mathring A$. The wavelength of the $L_\alpha$ line in the $X$-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:

In an $X-$ray experiment,what do $K_{\alpha}$ and $K_{\beta}$ denote?

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

When $0.50\; \mathring{A}$ $X$-rays strike a material,the photoelectrons from the $K$-shell are observed to move in a circle of radius $23\; mm$ in a magnetic field of $2 \times 10^{-2}\; T$ acting perpendicularly to the direction of emission of photoelectrons. The binding energy of $K$-shell electrons is $..........\; keV$.