X – rays are emitted by a tube containing the element Niobium (Z=41) as anti cathode,  and the $K_{\propto }$ X –rays are allowed to be incident on an unknown gas containing hydrogen  like ions it is found that the X – rays cause the emission of photoelectrons with an  energy of 2.7 ke V from these ions. Choose the correct options. (The intensity of X – rays is $100mW/m^2$ and 1% of the X – rays cause ionization)

The wavelength of the $K_{\alpha }$ radiation from Nb is $\lambda =\frac{4hc}{3R}\frac{1}{{(Z-1)}^2}=\frac{4hc}{3R}\frac{1}{{40}^2}$

$E_{K_{\alpha }}=16.3\text{keV\hspace{1em}}\therefore Z_{\text{gas }}^2=1000,Z_{\text{gas }}=32$

If the moment of the electrons is doubled, then the K.E. is increased by 4 times

$\therefore$The minimum energy of the $X$-rays is $(16.3+3\times 2.7)\text{keV}=24.5\text{keV}$

Since, $E_{K_{\alpha }}\propto {(Z-1)}^2$

${\left(\frac{Z-1}{40}\right)}^2=\left(\frac{24.5}{16.3}\right)Z=50$

The no. of $X$ -ray photons incident $=100 \text{mW}/{\text{m}}^2=\frac{100\times {10}^{-3}\times {10}^{19}}{1.6\times 24.5\times {10}^3}{\text{ per m }}^2$

No. of ions produced in 1 sec $=\frac{100\times {10}^{-3}\times {10}^{-2}\times {10}^{19}}{1.6\times 24.5\times {10}^3\times 3\times {10}^8}{\text{ per m}}^3\approx 850{\text{ per m}}^3$