MCQ
Let ${\lambda _\alpha }$, ${\lambda _\beta }$ and ${\lambda '_\alpha }$ denote the wavelengths of the $X-$ rays of the ${K_\alpha },\,{K_\beta }$ and ${L_\alpha }$ lines in the characteristic $X-$ rays for a metal
  • A
    ${\lambda _\alpha } > {\lambda '_\alpha } > {\lambda _\beta }$
  • B
    ${\lambda '_\alpha } > {\lambda _\beta } > {\lambda _\alpha }$
  • $\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}$
  • D
    $\frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{\lambda _\beta }}} = \frac{1}{{{{\lambda '}_\alpha }}}$

Answer

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 }}}$

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