The ionization energies of Cobalt $(Co)$,Copper $(Cu)$,and Molybdenum $(Mo)$ are $7.8 \, keV$,$9.0 \, keV$,and $20.1 \, keV$ respectively. If any of these metals are used as a target in an $X$-ray tube operating at $15 \, kV$,then:

The minimum frequency $\nu$ of continuous $X$-rays is related to the applied potential difference $V$ as

Fraunhofer lines of the solar system are an example of

Hard $X$-rays for the study of fractures in bones should have a minimum wavelength of $10^{-11} \, m$. The accelerating voltage for electrons in the $X$-ray machine should be:

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.

If the frequency of the $K_\alpha$ line of an element with $Z = 64$ is $\nu_\alpha$ and the frequency of the $K_\alpha$ line of an element with $Z = 80$ is $\nu'_\alpha$,then the ratio of their frequencies is: