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What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K?Assume that the molecule is moving with the root-mean square speed of molecules at this this temperature.(Atomic mass of nitrogen-14.0076 u) |
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Answer» Solution :Mass of nitrogen atom, `m=2xx14xx1.66xx10^(-27)=46.48xx10^(-27)kg` T=300 K `K_(B)`=Boltzmaan.s constant `=1.38xx10^(-23)Jmol^(-1)K^(-1)` `h=6.63xx10^(-34)Js` `implies` AVERAGE kinetic energy `lt(1)/(2)MV^(2)gt=(3)/(2)k_(B)T` `therefore v=sqrt((3K_(B)T)/(m))` `therefore` Momentum p=mv=`sqrt(3mk_(g)T)` `therefore` de-Broglie wavelength, `lambda=(h)/(p)` `therefore lambda=(h)/(sqrt(3mk_(B)T))` `=(6.63xx10^(-34))/(sqrt(3xx46.49xx10^(-27)xx1.38xx10^(-23)xx300))` `=(6.63xx10^(-34))/(sqrt(57728.16xx10^(-50)))` `=(6.63xx10^(-34))/(240.266xx10^(-25))` `=0.02759xx10^(-9)m` `~~0.028nm` |
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