Correct option: C.$\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}$
c
(c) According to the energy diagram of $X-$ ray spectra
$\because \Delta E = \frac{{hc}}{\lambda }$
$ \Rightarrow \lambda \propto \frac{1}{{\Delta E}}$
(${\Delta}E = $Energy radiated when e-jumps from, higher energy orbit to lower energy orbit)
${(\Delta E)_{{k_\beta }}} > {(\Delta E)_{{k_\alpha }}} > {(\Delta E)_{{L_\alpha }}}$$\therefore {\lambda '_\alpha } > {\lambda _\alpha } > {\lambda _\beta }$
Also ${(\Delta E)_{{k_\beta }}} = {(\Delta E)_{{k_\alpha }}} + {(\Delta E)_{{L_\alpha }}}$
$ \Rightarrow \frac{{hc}}{{{\lambda _\beta }}} = \frac{{hc}}{{{\lambda _\alpha }}} + \frac{{hc}}{{{{\lambda '}_\alpha }}}$
$ \Rightarrow \frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}$