Consider a metallic cube of edge length L. It | vedclass

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

Question

(B) Given the resistance formula $R = \frac{m_e v}{n e^2 L^2}$.From the de-Broglie wavelength relation,$\lambda = \frac{h}{p} = \frac{h}{m_e v}$.Given

Vedclass · Accepted Answer

$10^4$

Vedclass · Answer

(B) Given the resistance formula $R = \frac{m_e v}{n e^2 L^2}$.From the de-Broglie wavelength relation,$\lambda = \frac{h}{p} = \frac{h}{m_e v}$.Given $\lambda = L$,we have $L = \frac{h}{m_e v}$,which implies $m_e v = \frac{h}{L}$.Substituting this into the resistance formula: $R = \frac{(h/L)}{n e^2 L^2} = \frac{h}{n e^2 L^3}$.Since $n = \frac{N}{V} = \frac{N}{L^3}$ (where $N$ is the number of electrons),we get $R = \frac{h}{N e^2}$.For maximum resistance,the number of electrons $N$ must be minimum. The minimum number of electrons in a metallic sample is $N = 1$.Thus,$R_{max} = \frac{h}{e^2} = \frac{6.63 	imes 10^{-34}}{(1.60 	imes 10^{-19})^2}$.$R_{max} = \frac{6.63 	imes 10^{-34}}{2.56 	imes 10^{-38}} \approx 2.59 	imes 10^4 \, \Omega$.This value is closest to $10^4 \, \Omega$.

Similar Questions

The electric field $E$,current density $J$ and conductivity $\sigma$ of a conductor are related as

In the given hollow cylindrical conductor,the current density is $J = \frac{J_0}{r^2}$,where $J_0$ is a constant and $r$ is the distance from the axis of the cylinder. If the radius of the inner surface is $a$ and the radius of the outer surface is $2a$,find the current passed through the conductor.

Current density in a cylindrical wire of radius $R$ is given as $J = \begin{cases} J_0 \left( \frac{x}{R} - 1 \right) & 0 \leqslant x < \frac{R}{2} \\ J_0 \frac{x}{R} & \frac{R}{2} \leqslant x \leqslant R \end{cases}$. The current flowing in the wire is:

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