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InterviewSolution
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What is the de Broglie wavelengthof a nitrogen molecule in air at 300 K ? Assume that 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 :`T =300K "" LAMBDA=?` `m=14.0076u=14.0076xx1.6xx10^(-27)kg ` `K_("max")=(3)/(2)KT=(3)/(2)xx1.38xx10^(-23)xx300=6.21xx10^(-21)J` `p=sqrt(2mK_("max"))=sqrt(2xx14.0076xx1.6xx10^(-27)xx6.21xx10^(-21))=1.7xx10^(-23)kg ms^(-1)` `lambda=(h)/(p)=(6.6xx10^(-34))/(1.7xx10^(-23))=3.9xx10^(-11)m` |
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