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If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is `10 A`, then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximatelyA. 0.3 ÅB. 3ÅC. 30ÅD. 10Å |
Answer» Correct Answer - A We have K.E.`=(P^(2))/(2m_(0)) =(hc)/(lambda_("min"))` `rArr P=sqrt((2hcm_(0))/(lambda_("min")))` Also `lambda_("de broglie")=h/p=sqrt((hlambda_("min"))/(2m_(0)C))` For `lambda_("min")=10Å: lambda_("de broglie")=0.3 Å 0` |
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