Question:
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:
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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A typical atomic size is the Bohr radius \(a_0 \approx 0.53\ \text{\AA}\); find which combination has dimension of length and this magnitude.
Updated On: Oct 31, 2026
- \(\frac{e^2}{4\pi\varepsilon_0 mc^2}\)
- \(\frac{4\pi\varepsilon_0 \hbar^2}{me^2}\)
- \(\frac{me^2}{4\pi\varepsilon_0 \hbar^2}\)
- \(\frac{4\pi\varepsilon_0 mc^2}{e^2}\)
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The Correct Option is B
Solution and Explanation
Atomic size is the Bohr radius \(a_0 = \frac{4\pi\varepsilon_0\hbar^2}{me^2}\).
Option (1) \(\frac{e^2}{4\pi\varepsilon_0 mc^2}\) is a length (classical electron radius) but equals about \(2.8\times10^{-15}\) m, far smaller than an atom.
Options (3) and (4) are the inverses of a length (\(\frac{me^2}{4\pi\varepsilon_0\hbar^2} = \frac1{a_0}\), and \(\frac{4\pi\varepsilon_0 mc^2}{e^2} = \frac{1}{r_e}\)), so their dimensions are wrong.
Option (2): \(\frac{4\pi\varepsilon_0\hbar^2}{me^2} = \frac{(1.11\times10^{-10})(1.05\times10^{-34})^2}{(9.1\times10^{-31})(1.6\times10^{-19})^2} \approx 5.3\times10^{-11}\) m, which is atomic size.
Answer: (2) \(\frac{4\pi\varepsilon_0 \hbar^2}{me^2}\)
Option (1) \(\frac{e^2}{4\pi\varepsilon_0 mc^2}\) is a length (classical electron radius) but equals about \(2.8\times10^{-15}\) m, far smaller than an atom.
Options (3) and (4) are the inverses of a length (\(\frac{me^2}{4\pi\varepsilon_0\hbar^2} = \frac1{a_0}\), and \(\frac{4\pi\varepsilon_0 mc^2}{e^2} = \frac{1}{r_e}\)), so their dimensions are wrong.
Option (2): \(\frac{4\pi\varepsilon_0\hbar^2}{me^2} = \frac{(1.11\times10^{-10})(1.05\times10^{-34})^2}{(9.1\times10^{-31})(1.6\times10^{-19})^2} \approx 5.3\times10^{-11}\) m, which is atomic size.
Answer: (2) \(\frac{4\pi\varepsilon_0 \hbar^2}{me^2}\)
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