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If ${lambda _{{K_alpha }}},{lambda _{{K_beta }}}$ and ${lambda _{{L_alpha }}}$ are the wavelengths of ${K_alpha },{K_beta }$ and ${L_alpha }$ lines, then

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${lambda _{{K_beta }}} = frac{{{lambda _{{K_alpha }}}{lambda _{{L_alpha }}}}}{{{lambda _{{K_alpha }}} + {lambda _{{L_alpha }}}}}$

${lambda _{{L_alpha }}} = frac{{{lambda _{{K_alpha }}}{lambda _{{K_beta }}}}}{{{lambda _{{K_alpha }}} + {lambda _{{K_beta }}}}}$

${lambda _{{L_alpha }}} = frac{{{lambda _{{K_alpha }}}{lambda _{{K_beta }}}}}{{{lambda _{{K_beta }}} - {lambda _{{K_alpha }}}}}$

${lambda _{{K_beta }}} = frac{{{lambda _{{K_alpha }}}{lambda _{{L_alpha }}}}}{{{lambda _{{K_alpha }}} - {lambda _{{L_alpha }}}}}$

answer is A.

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Detailed Solution

${E_{K_beta }} = {E_{K_alpha }} + {E_{L_alpha }},;,,therefore frac{{hc}}{{{lambda _{K_beta }}}} = frac{{hc}}{{{lambda _{K_alpha }}}} + frac{{hc}}{{{lambda _{L_alpha }}}}$

$therefore {lambda _{K_beta }} = frac{{{lambda _{K_alpha }}{lambda _{L_alpha }}}}{{{lambda _{K_alpha }} + {lambda _{L_alpha }}}}$

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