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 is roughly equal to the known size of an atom (~ 10–10m).
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 is roughly equal to the known size of an atom (~ 10–10m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, m, and c. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in a non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

Ans:

(a)

We are given,

The charge on an electron, e = 1.6 x 10-19 C

Mass of an electron, me = 9.1 x 10-31 kg

Speed of the light, c = 3 x 108 m/s

The equation comprising of above quantities is given by the relation –

    \[\frac{{{e}^{2}}}{4\pi {{\in }_{0}}{{m}_{e}}{{c}^{2}}}\]

Where,

    \[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

ε0 is the permittivity of free space

The numerical value of the quantity is

    \[=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{9.1\times {{10}^{-31}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}}=2.81\times {{10}^{-15}}m\]

The numerical value of the equation taken is much lesser than the size of the atom.

(b)

Charge on an electron is given by e = 1.6 x 10-19 C

Mass of the electron is me = 9.1 x 10-31 kg

Planck’s constant is given by h = 6.63 x 10-34  Js

Now, writing a quantity involving all these –

    \[\frac{4\pi {{\in }_{0}}{{\left[ \frac{h}{2\pi } \right]}^{2}}}{{{m}_{e}}{{e}^{2}}}\]

Where, ε0 is the Permittivity of free space and

    \[\frac{1}{4\pi {{\in }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}\]

So, the numerical value of above equation becomes –

    \[\frac{4\pi {{\in }_{0}}{{\left[ \frac{h}{2\pi } \right]}^{2}}}{{{m}_{e}}{{e}^{2}}}=\frac{1}{9\times {{10}^{9}}}\times \frac{{{\left( \frac{6.63\times {{10}^{-34}}}{2\times 3.14} \right)}^{2}}}{9.1\times {{10}^{-31}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}\]

    \[=0.53\times {{10}^{-10}}m\]

Therefore, it can be seen that the numerical value of the quantity is of the order of the atomic size.