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Find the typical de-Broglie wavelength associated with a He atom in helium gas at room temperature (27^(@)C) and 1atm pressure, and compare it with the mean separation between two atoms under these conditions. |
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Answer» SOLUTION :Mass of He atom `=("ATOMIC mass of He")/("Avogadro.s number")=(4xx10^(-3))/(6xx10^(26))kg`[Taking `N_(A)=6xx10^(26)kg" "mol^(-1)`] `lamda_("de-broglie")=(h)/(p)=(h)/(sqrt(3mk_(B)T))""[becausep=mv_(rms)=msqrt((3k_(B)T)/(m))=sqrt(3mk_(B)T)]` `=(6.63xx10^(-34))/(sqrt(3xx.67xx10^(-27)xx1.38xx10^(-23)xx300))m=(6.63xx10^(-10))/(sqrt(82.84))m=7.3xx10^(-11)m` For one MOLE of a gas, `PV=RT=N_(A).k_(B)T` MEAN separation between atoms is `r=((V)/(N_(A)))^(1//3)=((k_(B)T)/(P))^(1//3)=((1.38xx10^(-23)xx300)/(1.01xx10^(5)))^(1//3)=3.4xx10^(-9)m` Therefore, `lamda_("de-Broglie") lt lt r`. |
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