Every atom contributes one free electron in c | vedclass

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(B) The number of atoms per unit volume $(n)$ is given by $n = \frac{	ext{Density} 	imes N_A}{	ext{Atomic Weight}}$.Given: Density $ho = 9 	imes

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$0.1 \ mm/s$

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(B) The number of atoms per unit volume $(n)$ is given by $n = \frac{	ext{Density} 	imes N_A}{	ext{Atomic Weight}}$.Given: Density $ho = 9 	imes 10^3 \ kg/m^3$,Atomic weight $M = 63 	imes 10^{-3} \ kg/mol$,Avogadro number $N_A = 6.023 	imes 10^{23} \ atoms/mol$.$n = \frac{9 	imes 10^3 	imes 6.023 	imes 10^{23}}{63 	imes 10^{-3}} \approx 8.6 	imes 10^{28} \ electrons/m^3$.The drift velocity $v_d$ is given by $v_d = \frac{I}{neA}$,where $I = 1.1 \ A$,$e = 1.6 	imes 10^{-19} \ C$,and $A = \pi r^2 = \pi (0.5 	imes 10^{-3})^2 \ m^2$.$v_d = \frac{1.1}{8.6 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 3.14 	imes 0.25 	imes 10^{-6}}$.$v_d \approx \frac{1.1}{1.08 	imes 10^4} \approx 1.01 	imes 10^{-4} \ m/s = 0.1 \ mm/s$.

Every atom contributes one free electron in copper. If $1.1 \ A$ current is flowing in a copper wire having a $1 \ mm$ diameter,then the drift velocity (approx.) will be (Density of copper $= 9 \times 10^3 \ kg \ m^{-3}$ and atomic weight $= 63$).

Similar Questions

Assertion: Free electrons always keep on moving in a conductor,even then no magnetic force acts on them in a magnetic field unless a current is passed through it.
Reason: The average velocity of free electrons is zero.

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Charge passing through a conductor of cross-section $0.3 \, m^2$ is given by $q = (3t^2 + 5t + 2) \, C$ where '$t$' is in seconds. The drift velocity at $t = 2 \, s$ is (Concentration of electrons in the conductor $= 2 \times 10^{25} \, m^{-3}$)

Derive the equation of mobility in terms of relaxation time $(\tau)$. Write its unit.

Drift speed of electrons,when $1.5 \, A$ of current flows in a copper wire of cross-section $5 \, mm^2$,is $v$. If the electron density in copper is $9 \times 10^{28} \, m^{-3}$,the value of $v$ in $mm/s$ is close to (Take charge of electron to be $1.6 \times 10^{-19} \, C$).

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