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For tungsten, atomic energy level of K, L & M are given 69.5 keV, 11.3 keV and 2.30 keV respectively. For obtaining charcteristic K_(beta)&K_(alpha)lines for tungsten, what should be the required minimum accelerating potential and lambda_("min") ? Also calculate lambda_(alpha) and lambda_(beta). |
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Answer» Solution :Required minimum accelerating POTENTIAL `=("ionisation energy")/(e)=69.5 kV` For this accelerating potential , `lambda_("min")=(HC)/(eV_("max"))=(12400)/(69.5 xx 10^(3))Å = 0.178Å` wavelength for `K_(alpha)` `lambda_(alpha)= (12400)/((69.5-11.3))Å=0.213Å` wavelength for `K_(beta)` `lambda_(beta)=(12400)/((69.5-230)xx10^(3))Å=0.184 Å` |
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