The quantum hall resistance R_H is a fundamen | vedclass

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Question

(b)

Let $R_H=k h^a e^b \quad \dots(i)$

As, $R=\frac{V}{I}$

$	herefore {\left[R_Hight] } =\left[ ML ^2 T ^{-3} A ^{-2}ight]$

$\Rightar

Vedclass · Accepted Answer

$\frac{h}{e^2}$

Vedclass · Answer

(b)

Let $R_H=k h^a e^b \quad \dots(i)$

As, $R=\frac{V}{I}$

$	herefore {\left[R_Hight] } =\left[ ML ^2 T ^{-3} A ^{-2}ight]$

$\Rightarrow h =E \cdot t$

$\Rightarrow {[h] }=\left[ ML ^2 T ^{-1}ight]$

$e=I \cdot t \Rightarrow[e]=[ A \cdot T ]$

Substituting above values in Eq. $(i)$, we have

${\left[ ML ^2 T ^{-3} A ^{-2}ight] }=k\left[ ML ^2 T ^{-1}ight]^e[ AT ]^b$

$=k\left[ M ^a L ^{2 a} T ^{-a+b} A ^bight]$

Equating dimensions, we get

$a=1 	ext { and } b=-2$

Hence, $\quad R_H=k\left(\frac{h}{e^2}ight)$

So, dimensions of hall resistance are same as that of $\frac{h}{e^2}$.

The quantum hall resistance $R_H$ is a fundamental constant with dimensions of resistance. If $h$ is Planck's constant and $e$ is the electron charge, then the dimension of $R_H$ is the same as

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