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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
$\frac{e^2}{h}$
$\frac{h}{e^2}$
$\frac{h^2}{e}$
$\frac{e}{h^2}$
Solution
(b)
Let $R_H=k h^a e^b \quad \dots(i)$
As, $R=\frac{V}{I}$
$\therefore {\left[R_H\right] } =\left[ ML ^2 T ^{-3} A ^{-2}\right]$
$\Rightarrow h =E \cdot t$
$\Rightarrow {[h] }=\left[ ML ^2 T ^{-1}\right]$
$e=I \cdot t \Rightarrow[e]=[ A \cdot T ]$
Substituting above values in Eq. $(i)$, we have
${\left[ ML ^2 T ^{-3} A ^{-2}\right] }=k\left[ ML ^2 T ^{-1}\right]^e[ AT ]^b$
$=k\left[ M ^a L ^{2 a} T ^{-a+b} A ^b\right]$
Equating dimensions, we get
$a=1 \text { and } b=-2$
Hence, $\quad R_H=k\left(\frac{h}{e^2}\right)$
So, dimensions of hall resistance are same as that of $\frac{h}{e^2}$.
Similar Questions
Match List $I$ with List $II$ and select the correct answer using the codes given below the lists :
| List $I$ | List $II$ |
| $P.$ Boltzmann constant | $1.$ $\left[ ML ^2 T ^{-1}\right]$ |
| $Q.$ Coefficient of viscosity | $2.$ $\left[ ML ^{-1} T ^{-1}\right]$ |
| $R.$ Planck constant | $3.$ $\left[ MLT ^{-3} K ^{-1}\right]$ |
| $S.$ Thermal conductivity | $4.$ $\left[ ML ^2 T ^{-2} K ^{-1}\right]$ |
Codes: $ \quad \quad P \quad Q \quad R \quad S $
