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At 10^(@)C tempreture ,de-Broglie wave length of atom is 0.4 Å.If temperatire of atom is increased by 30^(@)C ,What will be change in de-Broglie of atom? |
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Answer» decreases `10^(2)Å` `T_(2)=40+273=313K` `lambda_(1)=0.4 Å` `lambda=(h)/(sqrt(2ME))` but,`E=(3)/(2)KT` so `lambda =(h)/(sqrt(3mKT))` `therefore lambda prop (1)/(sqrt(T))` `therefore (lambda_(2))/(lambda_(1))=sqrt(T_(1)/(T_(2)))` `therefore (lambda_(2))/(0.4xx10^(-10))=sqrt((283)/(313))` `therefore (lambda_(2))/(0.4xx10^(-10))=0.951` `therefore lambda_(2)=0.951xx0.4xx10^(-10)` `therefore lambda_(2)0.38 Å` `therefore "wavelength" =lambda_(2)-lambda_(1)` |
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