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InterviewSolution
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A copper wire fo cross-sectional area `3.00xx10^(-6) m^2` carries a current 10.0A. a. Find the drift speed of the electrons in the wire. Assume that each copper atom contributes one free electron to the body of material. b. Find the average time between collisions for electrons in the copper at `20^@C.` The density of copper si `8.95 gcm^(-1)`, molar mass of copper is `63.5 gmol^(-1)`, Avogadro number is `6.02x10^(23)` electrons per mol and resistivity of copper is 1.7 xx10^(-8) Omegam.` |
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Answer» The volume occupied by 63.5 g of copper is `V = (M)/(rho) = (63.5)/(8.95) = 7.09 cm^3 mol^(-1)` As each copper atom contributes one free electron to the body of the meterial, the density of freee electrons is `n = (6.02xx10^(23))/(7.09xx10^(-6)) = 8.48xx10^(28)` electrons per cubic meter The drift speed is `v_d = (I)/("ne"A)` `=(10.0)/(8.48xx10^(28)xx1.60xx10^(-19)xx3xx10^(-6)) = 2.46xx10^(-4)ms^(-1)` b. Average time between collision of electrons is `tau = (m_e)/("ne"^2rho) = (9.10xx10^(31))/(8.48xx10^(28)xx(1.6xx10^(19))^(2)xx1.7xx10^(-8))` `=2.5xx10^(-14)S` |
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