In an X-ray tube the accelerating voltage is 20 kV. Two targets A and B are used one by one. For ‘A’ the wavelength of the $K_{\alpha }$  line is 62 pm. For ‘B’ the wavelength of the  $L_{\alpha }$ line is 124pm. The energy of the ‘B’ ion with vacancy in ‘M’ shell is 5.5 keV higher than the atom of B. $[Take h c = 12400 e V{A}^{°}]$

(A)  ${\lambda }_{\min }=\frac{hc}{K.Eof{e}^{-}}=\frac{12400}{20\times {10}^3}\stackrel{\circ }{\text{A}}=0.62\stackrel{\circ }{\text{A}}=62pm.$
(B) Since ${\lambda }_{\min }=62pm,K_{\alpha }$  from A will not be obtained.
(C) L–photons can be emitted if electron from L–shell can be removed the energy  required to remove L–shell electrons.
 $=55.5KeV+\frac{12400}{124\times {10}^{-2}}eV=15.5KeV.$
As the energy of encoming electron is 20 keV > 15.5 keV, the L–shell electron can be  removed. Hence, L photon can be obtained.
The minimum wavelength will correspond to the transimition from  $\infty$ to L–shell.
 $\lambda =\frac{12400}{15.5\times {10}^3}A=0.8\stackrel{\circ }{\text{A}}$