From the following, the quantity (constructed from the basic constants of nature), that has the dimensions,…

From the following, the quantity (constructed from the basic constants of nature), that has the dimensions, as well as correct order of magnitude, vis-a-vis typical atomic size, is:
  1. $\frac{e^2}{4 \pi \varepsilon_0 m c^2}$
  2. $\frac{4 \pi \varepsilon_0 e^2}{m e^2}$
  3. $\frac{m e^2}{4 \pi \varepsilon_0 b^2}$
  4. $\frac{4 \pi \varepsilon_0 m c^2}{e^2}$

Solution

As $\mathrm{E}=\mathrm{mc}^2$ also $\mathrm{E}=\mathrm{F} \cdot \mathrm{s}=\frac{\mathrm{kq} \cdot \mathrm{q}}{\mathrm{r}^2} \cdot \mathrm{r}$ Therefore dimensionally $\begin{aligned} & \mathrm{mc}^2=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}^2}{\mathrm{r}} \\ & \Rightarrow \mathrm{r}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{mc}^2} \end{aligned}$

Asked in: JEE Main 2013 (09 Apr Online)

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