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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 2 xx 10^(-10) m. |
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Answer» Solution :`T=27+273=300 K,K_(B)=1.38xx10^(-23)Jmo,^(-1)K^(-1)` `m=9.1xx10^(-31)kg,h=6.63xx10^(-34)JS` DISTANCE between two electron `r=2xx10^(-10)m` `implies` de-Broglie wavelength of electron, `lambda=(h)/(sqrt(3mk_(B)T))` `=(6.63xx10^(-34))/(sqrt(3xx9.1xx10^(-31)xx1.38xx10^(-28)xx300))` `(6.63xx10^(-34))/(sqrt(11302.3xx10^(-54)))` `=(6.63xx10^(-34))/(106.31xx10^(-27))` `=0.06236xx10^(-7)` `~~6.2xx10^(-9)`m `implies` Average distance between two electron in metal, `r=2xx10^(-10)`m `therefore (lambda)/(r)=(6.2xx10^(-9))/(2xx10^(-10))=31` Hence average distance between two electron in atom is 31 times de-Broglie wavelength of electron. |
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