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A copper wire of cross sectional area 0.5 mm^2 carries a currentsof 1A . Assume that each copper atom contributes one free electron . The density of copper is 9,000 kg//m^3 and atomic mass is 63.5 u. if v_d is drift speed of electrons. Now assume that one copper atom is kept at temperature 300 K and it develops thermal motion. Let v_("rms") be the root mean square speed of copper atom at given temperature . find the ratio v_(rms)//v_d ( HInt: Treat copper atom like a mono-atomic gas particle for the calculation of v_(rms)) |
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Answer» Solution :Cross SECTIONAL area of wire `A=0.5 mm^2 =0.5 xx10^(-6)m^2` Current I=1A Density of copper `d=9,000 kg //m^3` Atomic mass of copper=63.5 u Drift velocity `v_d=?` Root mean square velocity `v_(rms)=?` Mass of `1 m ^3` of copper=9,000 kg Number of atoms in 63.5 g of Cu`=6.023 xx 10^(23)` Number of atoms in 1g of `Cu =(6.023xx10^(23))/(63.5)` Number of atoms in `9,000 xx10^3 g of Cu`, `n=(6.023xx10^(23))/(63.5)xx9,000xx10^3` `rArr n=85.3 xx10^(27)` atoms As `I="neA"v_d` `I/("neA")=v_d` `rArr 1/(85.3xx10^(27)xx1.6xx10^(-19)xx0.5xx10^(-6))=v_d` `therefore v_d =0.015xx10^(-2)m//s` `v_(rms)=sqrt((3kT)/m)` Mass of copper atom ,m `=(63.5)/(6.023xx10^(23))g` `=(63.5xx10^(-3))/(6.023xx10^(23))kg` k= BOLTZMANN constant `=1.38xx10^(-23) m^2 kg s^(-2)K^(-1)` `v_(rms)=sqrt((3xx1.38xx10^(-23)xx300xx6.023xx10^(23))/(63.5xx10^(-3)))` `=3.43xx10^2m s^1 rArr 343 ms^(-1)` `therefore v_(rms)//v_d=0.015xx10^(-2)//343=4.37xx10^(-5)` |
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