A metal has 9 times 10^{28} conduction electr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The resistivity $ho$ is given by $ho = \frac{m}{ne^2	au}$, where $	au$ is the relaxation time. Also, the mean free path $\lambda$ is given b

Vedclass · Accepted Answer

$62.5$

Vedclass · Answer

(D) The resistivity $ho$ is given by $ho = \frac{m}{ne^2	au}$, where $	au$ is the relaxation time. Also, the mean free path $\lambda$ is given by $\lambda = v_d 	au$, where $v_d$ is the drift speed. From the resistivity formula, $	au = \frac{m}{ne^2ho}$. Substituting this into the mean free path formula: $\lambda = v_d \left( \frac{m}{ne^2ho} ight) = \frac{m v_d}{ne^2ho}$. Given: $m = 9 	imes 10^{-31} \ kg$, $v_d = 1.6 	imes 10^6 \ m/s$, $n = 9 	imes 10^{28} \ m^{-3}$, $e = 1.6 	imes 10^{-19} \ C$, and $ho = 1 	imes 10^{-8} \ \Omega \cdot m$. $\lambda = \frac{(9 	imes 10^{-31}) 	imes (1.6 	imes 10^6)}{(9 	imes 10^{28}) 	imes (1.6 	imes 10^{-19})^2 	imes (1 	imes 10^{-8})}$. $\lambda = \frac{14.4 	imes 10^{-25}}{9 	imes 10^{28} 	imes 2.56 	imes 10^{-38} 	imes 10^{-8}} = \frac{14.4 	imes 10^{-25}}{23.04 	imes 10^{-18}} = 0.625 	imes 10^{-7} \ m = 62.5 \ nm$.

Similar Questions

The quantities that do not change when a resistor connected to a battery is heated due to the current are:
$(A)$ drift speed
$(B)$ resistivity
$(C)$ resistance
$(D)$ number of free electrons

The area of cross-section of a copper wire is $4 \times 10^{-7} \,m^2$ and the number of free electrons per cubic metre in copper is $8 \times 10^{28}$. If the wire carries a current of $6.4 \,A$,then the drift velocity of the electrons (in $10^{-3} \,m \,s^{-1}$) is:

The drift velocity of electrons in a current-carrying wire of radius $r$ is $v$. If we want a drift velocity of $2v$ in a wire of the same material but with half the radius,what current must be passed through it?

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 $= 63$).

The drift velocity of electrons in a conducting wire connected to a cell is $V_{d}$. If the length of the wire is doubled and the area of cross-section is halved,then the drift velocity of electrons becomes:

Vedclass Test Series

Exam Paper Generator

Online Exam Module