12.14 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^{-10}m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, 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 non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to 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.

Charge on an electron, e= $1.6\times {10}^{-19}C$

Mass of an electron, ${\mathrm{m}}_{\mathrm{e}}=9.1\times {10}^{-31}kg$

Speed of light, $c=3\times {10}^{8}m/s$

Let us take a quantity involving the given quantities as $\left(\frac{{e}^{2}}{4\mathrm{\pi}{\in}_{0}{m}_{e}{c}^{2}}\right).$

where,

${\in}_{0}$ = Permittivity of free space and

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

The numerical value of the taken quantity will be:

\(\frac{e^{2}}{4 \pi \in_o{m}_e{c^{2}}}=9\times 10^{9}\times \frac{1.6\times 10^{-19}}{9.1\times 10^{-31}\times \left ( 3\times 10^{8} \right )^{2}}=2.81\times 10^{-15} m\)

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom

**(b)**

Step 2: Find the combination involving Planck's constant.

Charge on an electron, \(e=1.6 \times 10^{-19} C\)

Mass of an electron, \(\mathrm{m}_{e}=9.1 \times 10^{-31} \mathrm{~kg}\)

Planck's constant, \(h=6.63 \times 10^{-34}~ J-sec\)

Let us construct a quantity involving the given quantities as:

\(\frac{e^{2}}{4 \pi \in_o{m}_e{c^{2}}}=9\times 10^{9}\times \frac{1.6\times 10^{-19}}{9.1\times 10^{-31}\times \left ( 3\times 10^{8} \right )^{2}}=2.81\times 10^{-15} m\)

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom

Step 2:

Charge on an electron, \(e=1.6 \times 10^{-19} C\)

Mass of an electron, \(\mathrm{m}_{e}=9.1 \times 10^{-31} \mathrm{~kg}\)

Planck's constant, \(h=6.63 \times 10^{-34}~ J-sec\)

Let us construct a quantity involving the given quantities as:

$4\mathrm{\pi}{\in}_{0}\times \frac{{\left({\displaystyle \frac{\mathrm{h}}{2\mathrm{\pi}}}\right)}^{2}}{{\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{9\times {10}^{9}}\times \frac{{\left({\displaystyle \frac{6.63\times {10}^{-34}}{2\times 3.14}}\right)}^{2}}{9.1\times {10}^{-31}\times {\left(1.6\times {10}^{-19}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.53\times {10}^{-10}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the quantity taken is of the order of the atomic size.