The wavelength of the ${K_\alpha}$ line for an element of atomic number $29$ is $\lambda$. Then the wavelength of the ${K_\alpha}$ line for an element of atomic number $15$ is (Take Moseley's constant $b = 1$ for both elements).

Hydrogen atom does not emit $X$-rays because

If ${\lambda _1}$ and ${\lambda _2}$ are the wavelengths of characteristic $X$-rays and gamma rays respectively,then the relation between them is

In a Coolidge tube experiment, the minimum wavelength of the continuous $X-$ray spectrum is equal to $66.3\, pm$. Then:

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 tube, electrons emitted from a filament (cathode) carrying current $I$ hit a target (anode) at a distance $d$ from the cathode. The target is kept at a potential $V$ higher than the cathode, resulting in the emission of continuous and characteristic $X$-rays. If the filament current $I$ is decreased to $I/2$, the potential difference $V$ is increased to $2V$, and the separation distance $d$ is reduced to $d/2$, then:
$(A)$ The cut-off wavelength will reduce to half, and the wavelengths of the characteristic $X$-rays will remain the same.
$(B)$ The cut-off wavelength as well as the wavelengths of the characteristic $X$-rays will remain the same.
$(C)$ The cut-off wavelength will reduce to half, and the intensities of all the $X$-rays will decrease.
$(D)$ The cut-off wavelength will become two times larger, and the intensity of all the $X$-rays will decrease.