The minimum wavelength of the $X$-rays produced at an accelerating potential $V$ is $\lambda$. If the accelerating potential is changed to $2V$, then the minimum wavelength would become:

The figure represents the observed intensity of $X-$rays emitted by an $X-$ray tube as a function of wavelength. The sharp peaks $A$ and $B$ denote:

Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on a target. The cut-off wavelength $\lambda_0$ is equal to [$h=$ Planck's constant,$c=$ velocity of light].

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 energy of the $L$ shell is $7200$ units and the energy of the $K$ shell is $1800$ units. Therefore,the energy of the $K_\alpha$ line is equal to ...... units.