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

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.

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:

The potential difference applied to an $X$-ray tube is increased. As a result,in the emitted radiation

$X$-rays of wavelength $0.1 \ \mathring A$ are allowed to fall on a metal and get scattered. The wavelength of the scattered radiation is $0.111 \ \mathring A$. If $h = 6.624 \times 10^{-34} \ J \cdot s$ and $m_0 = 9 \times 10^{-31} \ kg$,then the direction of the scattered photons will be:

$X$-rays are produced by accelerating electrons through a potential difference $V$ and letting them strike a metal target of atomic number $Z$. The highest frequency of the $X$-rays produced is proportional to: