A) \[\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}\,m{{c}^{2}}}\]
B) \[\frac{4\pi {{\varepsilon }_{0}}\,{{b}^{2}}}{m{{e}^{2}}}\]
C) \[\frac{m{{c}^{2}}}{4\pi {{\varepsilon }_{0}}\,{{b}^{2}}}\]
D) \[\frac{4\pi {{\varepsilon }_{0}}\,m{{c}^{2}}}{{{e}^{2}}}\]
Correct Answer: B
Solution :
The typical atomic size given by \[\frac{4\pi {{\varepsilon }_{0}}{{h}^{2}}}{m{{e}^{2}}}\]
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