Question

The dimensions of the quantity namely $\frac{{\mu }_{0}c{e}^2}{2\hslash },$ is where ${\mu }_0-$ permeability of free space $c-$ velocity of light, $e-$ electronic charge and $\hslash =\frac{h}{2\pi },h$ being Planck's constant

• A. $\left[M^{0}LT\right]$
• B. $\left[M^0{L}^{0}T\right]$
• C. $\left[M^0{L}^0{T}^0\right]$
• D. $\left[M^{-2}{L}^{-1}{T}^{-3}\right]$

Answer 1

Answer: (C)

The quantity is $\frac{{\mu }_{0}c{e}^2}{2\hslash }=\frac{\pi {\mu }_{0}c{e}^2}{h}$
$\because c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$
or $c^2=\frac{1}{{\mu }_0{\epsilon }_0}$
$\therefore c{\mu }_0=\frac{1}{c{\epsilon }_0}$
$\therefore \frac{\pi {\mu }_{0}c{e}^2}{h}=\frac{\pi e^2}{c{\epsilon }_{0}h}$
$\because F=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$
or $\frac{e^2}{{\epsilon }_0}=F{r}^{2}4\pi$
$\therefore \frac{\pi e^2}{c{\epsilon }_{0}h}=\frac{F{r}^2}{ch}\times 4{\pi }^2$
The dimensions of
$\frac{F{r}^2}{ch}=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[L{T}^{-1}\right]\left[M{L}^2{T}^{-1}\right]}$
$=\frac{\left[M{L}^3{T}^{-2}\right]}{\left[M{L}^3{T}^{-2}\right]}$
$=\left[M^0{L}^0{T}^0\right]$