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Compute the typical de-Broglie wavelength of an electron in a metal at 27^(@)C and compare it with the mean separation between two electrons in a metal which is given to be about 2xx10^(-10)m. |
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Answer» Solution :Electrons in metal behave like ann electon gas. For electron `lamda_("de-Broglie")=(h)/(p)=(h)/(sqrt(3mk_(B)T))=(6.63xx10^(-34))/(sqrt(3xx.911xx10^(-31)xx1.38xx10^(-23)xx300))=6.2xx10^(-9)m` Given `r=2xx10^(-10)m` `THEREFORE (lamda_("de-Broglie"))/(r)=(6.2xx10^(-9))/(2xx10^(-10))=31` Therefore, `lamda_("de-Broglie")gtgt`mean separation between electrons in a metal. |
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