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Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27^(@)C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions. |
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Answer» Solution :Atomic weight of He `=4g=4xx10^(-3)Kg` Pressure P=1 atm.=`1.01xx10^(5) N//m^(2)` Boltzmaan.s constant, `k_(B)=1.38xx10^(-23)Jmol^(-1)K^(-1)` T=27+273 =300 K,`N_(0)=6xx10^(23)` `implies` Mass of given atom, `m=("atomic weight of He")/("No. of Avogardro")` `therefore m=(4xx10^(-3))/(6xx10^((23))` `=(2)/(3)xx10^(-26) kg` `implies` At standard temperature average kinetic energy of He atom, `(1)/(2)mv^(2)=(3)/(2)K_(B)T` `therefore m^(2)v^(2)=3mk_(B)T` `therefore p^(2)=3mk_(B)T`. `therefore p=sqrt(3mmk_(B)T)` Now ,de-Broglie wavelength, `lambda=(h)/(p)=(h)/(sqrt(3mk_(B)T))` `therefore lambda=(6,63xx10^(-34))/(sqrt(3xx(2)/(3)xx10^(-26)xx1.38xx10^(-23)xx300))` `therefore lambda=(6.63xx10^(-34))/(sqrt(828xx10^(-49)))` `=(6.63xx10^(-34))/(90.99xx10^(-25))` `therefore lambda=0.0728xx10^(-9)` `lambda ~~0.73xx10^(-10)m` `implies` Equation of state of 1 mole gas, PV=RT `therefore PV=N_(A)k_(B)T` `[because k_(B)=(R)/(N_(A))impliesR=N_(A)k_(B)]` `therefore (V)/(N_(A))=(k_(B)T)/(P)` `"Average distance "r=[("Molar volume")/("Avogardo number")]^((1)/(3))` `(V)/(N_(A))` `[(k_(B)T)/(P)]^((1)/(3))` `[(1.38xx10^(-23)xx300)/(1.01xx10^(5))]^((1)/(3))` `r=[(138xx30)/(101)xx10^(-27)]^((1)/(3))` log r=`[40.99]^((1)/(3))xx10^(-9)` `logr=(1)/(3)log[40.99]xx10^(-9)` `log=(1)/(3)[1.6127]xx10^(-9)` `logr=0.5378xx10^(-9)` Antilog`r=3.449xx10^(-9)` `therefore r~~3.4xx10^(-9)m` Here `gt lambda` hence average distance between two atom is very large compared to de-Broglie wavelength. |
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