Question

Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that are roughly equal to the known size of an atom (\[{{10}^{-10}}\]m).
Construct a quantity with the dimensions of length from the fundamental constants e, \[{{m}_{e}}\], and C. determine its numerical value.

Answer 1

Hint: According to coulomb’s law, the force between hydrogen nucleus and electron can be calculated with the following formula.
\[\begin{align}
  & F=\dfrac {1}{4\pi {{\varepsilon }_{0}}} \times \dfrac {{{e}^{2}}}{{{r}^{2}}} \\
 & r=\dfrac{1}{4\pi {{\varepsilon }_{0}}} \times \dfrac{{{e}^{2}}}{F.r}\to (1) \\
\end{align}\]
Where \[{{\varepsilon }_{0}}\]= permittivity of free space
r = distance
F =Force
e = charge of the electron.

Complete step by step answer:
We know that the charge of the electron, e = \[1.6\times {{10}^{-19}}C\]
Mass of the electron, \[{{m}_{e}}\]= \[9.1\times {{10}^{-31}}kg\]
Speed of the light, C =\[3\times {{10}^{8}}m/s\]
We know that the multiplication of force and distance is the amount work and it is equal to \[m{{c}^{2}}\]
\[\begin{align}
  & \text{force }\!\!\times\!\!\text{ distance}=m{{c}^{2}} \\
 & \text{distance=}\dfrac{\text{m}{{\text{c}}^{\text{2}}}}{\text{force}}\to (2) \\
\end{align}\]
Now we can substitute equation (2) in (1)

\[r=\dfrac{1}{4\pi {{\varepsilon }_{0}}}.\dfrac{{{e}^{2}}}{{{m}_{e}}{{C}^{2}}}\to (3)\]
Where \[{{\varepsilon }_{0}}\]= permittivity of free space
r = distance
F = Force
e = charge of the electron
C = velocity of light
\[\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times 109N{{m}^{2}}{{C}^{-2}}\]
Now substitute all the known values in equation 3.
 \[\begin{align}
  & =9\times {{10}^{9}}\times \dfrac{{{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{9.1\times {{10}^{-31}}{{\left( 3\times {{10}^{8}} \right)}^{2}}} \\
 & =2.81\times {{10}^{-15}}m \\
\end{align}\]

Note: The numerical quantity obtained with the dimensions of length from the fundamental constants e, \[{{m}_{e}}\], and C is smaller than the size of the atom.
The size of the Bohr’s model of hydrogen atom \[=0.529\times {{10}^{-10}}\overset{o}{\mathop{\text{A}}}\,\].