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What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume the molecule is moving with the root-mean-square speed of molecules at this temperature. [Atomic mass of nitrogen=14.0076 u]. |
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Answer» Solution :We know that `1u=1.66xx10^(-27)kg` and nitrogen is DIATOMIC gas. `THEREFORE` MASS of nitrogen molecule`=2xx14.0076xx1.66xx10^(-27)kg=4.649xx10^(-26)kg` Temperature of nitrogen molecule `T=300K` `therefore rms` speed of `N_(2)` molecule `v_(rms)=sqrt((3k_(B)T)/(m))` `therefore` de-Broglie wavelength `lamda=(h)/(mv_(rms))=(h)/(sqrt(3mk_(B)T))=(6.63xx10^(-34))/(sqrt(3xx4.649xx10^(-26)xx1.38xx10^(-23)xx300))` `=2.8xx10^(-11) m or 0.028nm`. |
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