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JEE Mains · Physics · STD 12 - 13. Nuclei
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
- A \(\frac{{{e^2}}}{{4\pi {\varepsilon _0}m{c^2}}}\)
- B \(\frac{{4\pi {\varepsilon _0}{e^2}}}{{m{e^2}}}\)
- C \(\frac{{m{e^2}}}{{4\pi {\varepsilon _0}{b^2}}}\)
- D \(\frac{{4\pi {\varepsilon _0}m{c^2}}}{{{e^2}}}\)
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 \(\mathrm{mc}^{2}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}}\)…
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