Question

The dimensions of the quantity namely $\frac{\mu_{0} c e^{2}}{2 \hbar},$ is where $\mu_{0}-$ permeability of free space $c-$ velocity of light, $e-$ electronic charge and $\hbar=\frac{h}{2 \pi}, h$ 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 $\frac{\mu_{0} c e^{2}}{2 \hbar}=\frac{\pi \mu_{0} c e^{2}}{h}$
$\because c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$
or $c^{2}=\frac{1}{\mu_{0} \varepsilon_{0}}$
$\therefore \quad c \mu_{0}=\frac{1}{c \varepsilon_{0}}$
$\therefore \frac{\pi \mu_{0} c e^{2}}{h}=\frac{\pi e^{2}}{c \varepsilon_{0} h}$
$\because F=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$
or $\frac{e^{2}}{\varepsilon_{0}}=F r^{2} 4 \pi$
$\therefore \frac{\pi e^{2}}{c \varepsilon_{0} h}=\frac{F r^{2}}{c h} \times 4 \pi^{2}$
The dimensions of
$\frac{F r^{2}}{c h}=\frac{\left[ MLT ^{-2}\right]\left[ L ^{2}\right]}{\left[ LT ^{-1}\right]\left[ ML ^{2} T ^{-1}\right]}$
$=\frac{\left[ ML ^{3} T ^{-2}\right]}{\left[ ML ^{3} T ^{-2}\right]}$
$=\left[ M ^{0} L ^{0} T ^{0}\right]$