Look up the values of the quantities in\[{{\bf{a}}_{\bf{B}}}{\bf{ = }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{{\bf{4}}{{\bf{\pi }}^{\bf{2}}}{{\bf{m}}_{\bf{e}}}{\bf{kq}}_{\bf{e}}^{\bf{2}}}}\], and verify that the Bohr radius a_B is 0.529 x 10^-10 m.

Consider the formula for the Bohr’s radius is given as:

\[{{\bf{a}}_{\bf{B}}}{\bf{ = }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{{\bf{4}}{{\bf{\pi }}^{\bf{2}}}{{\bf{m}}_{\bf{e}}}{\bf{k}}{{\bf{q}}_{\bf{e}}}^{\bf{2}}}}\]

Here, a_BBohr's radius, his the Planck's constant, m_e is the mass of electron, k is the electrostatic constant, and \({q_e}\) is the charge of electron

Consider the values are as follows:

\[\begin{array}{l}h = 6.626 \times {10^{ - 34}}\;\;{\rm{J}} \cdot {\rm{s}}\\{m_e} = 9.11 \times {10^{ - 31}}\;\;{\rm{kg}}\\{q_e} = 1.60 \times {10^{ - 19}}\;{\rm{C}}\\k = 9 \times {10^9}\;\frac{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}\end{array}\]

As a result, Bohr's radius is calculated as,

\[\begin{array}{l}{a_B} = \frac{{{{\left( {6.626 \times {{10}^{ - 34}}\;\;{\rm{J}} \cdot {\rm{s}}} \right)}^{\rm{2}}}}}{{{\rm{4}}{{\rm{\pi }}^{\rm{2}}}\left( {9.11 \times {{10}^{ - 31}}\;\;{\rm{kg}}} \right)\left( {9 \times {{10}^9}\;\frac{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}} \right){{\left( {1.60 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^{\rm{2}}}}}\\{a_B} = 5.2984 \times {10^{ - 11}}\;{\rm{m}}\\{a_B} = 0.52984 \times {10^{ - 10}}\;\;{\rm{m}}\end{array}\]

Hence, it is proved that Bohr radiusa_B is 0.529 x 10^-9 m.