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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2013
  • Medium
  • MCQ

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

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Answer & Solution

Correct Answer

(A) \(\frac{{{e^2}}}{{4\pi {\varepsilon _0}m{c^2}}}\)

Step-by-step Solution

Detailed explanation

As \(\mathrm{E}=\mathrm{mc}^{2}\)

also \(\mathrm{E}=\mathrm{F} \cdot \mathrm{s}=\frac{\mathrm{k} q \cdot \mathrm{q}}{\mathrm{r}^{2}} \cdot \mathrm{r}\)

Therefore dimensionally

\(\mathrm{mc}^{2}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}}\)

\(\Rightarrow r=\frac{e^{2}}{4 \pi \epsilon_{0} m c^{2}}\)

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