Question

If $e$ is the electronic charge, $c$ is the speed of light in free space and $h$ is Planck's constant, the quantity $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{hc}$ has dimensions of:

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

Answer 1

The correct option is B

$\left[M^0{L}^0{T}^0\right]$

Step 1: Given data

$e\to$electronic charge

$c\to$speed of light in a vacuum

$h\to$Planck's constant

We have to find the dimensions of $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{hc}$

Step 2: Formula used

We know,

Force, $F=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{r^2}orF{r}^2=\frac{e^2}{4{\mathrm{\pi \epsilon }}_0}$

Energy, $E=\frac{hc}{\lambda }orE\lambda =hc$

$$\begin{gathered} \therefore dim\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{hc}\right)=dim\left(\frac{F}{E}\frac{r^2}{\lambda }\right) \\ \Rightarrow dim\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{hc}\right)=\frac{\left[ML{T}^{-2}\right]}{\left[M{L}^2{T}^{-2}\right]}\frac{\left[L^2\right]}{\left[L\right]} \\ \Rightarrow dim\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{hc}\right)=\left[M^0{L}^0{T}^0\right] \end{gathered}$$

We get, the dimensions of $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{hc}$ is $\left[M^0{L}^0{T}^0\right]$

Hence, option B is correct.