हाइड्रोजन परमाणु में इलेक्ट्रॉन एवं प्रोटॉन के मध्य गुरुत्वाकर्षण, कूलॉम-आकर्षण से लगभग $10^{-40}$ के गुणक से कम है। इस तथ्य को देखने का एक वैकल्पिक उपाय यह है कि यदि इलेक्ट्रॉन एवं प्रोटॉन गुरुत्वाकर्षण द्वारा आबद्ध हों तो किसी हाइड्रोजन परमाणु में प्रथम बोर कक्षा की त्रिज्या का अनुमान लगाइए। आप मनोरंजक उत्तर पाएँगे।

Radius of the first Bohr orbit is given by the relation,

$r_{1}=\frac{4 \pi \epsilon_{0}\left(\frac{h}{2 \pi}\right)^{2}}{m_{e} e^{2}}\dots(i)$

Where,

$\epsilon_{0}=$ Permittivity of free space

$h=$ Planck's constant $=6.63 \times 10^{-34}\, Js$

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

$e=$ Charge of an electron $=1.9 \times 10^{-19}\, C$

$m_{p}=$ Mass of a proton $=1.67 \times 10^{-27} \,kg$

$r=$ Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as

$F_{c}=\frac{e^{2}}{4 \pi \epsilon_{0} r^{2}}\dots(ii)$

Gravitational force of attraction between an electron and a proton is given as

$F_{G}=\frac{G m_{p} m_{e}}{r^{2}}\dots(iii)$

Where,

$G=$ Gravitational constant $=6.67 \times 10^{-11} \,Nm ^{2} / kg ^{2}$

The electrostatic (Coulomb) force and the gravitational force between an electron ahd a proton are equal, then we can write

$\therefore F_{G}=F_{C}$

$\frac{G m_{p} m_{e}}{r^{2}}=\frac{e^{2}}{4 \pi \epsilon_{0} r^{2}}$

$\therefore \frac{e^{2}}{4 \pi \epsilon_{0}}=G m_{p} m_{e}\dots(iv)$

Putting the value of equation $(iv)$ in equation $(i),$ we get

$r_{1}=\frac{\left(\frac{h}{2 \pi}\right)^{2}}{G m_{p} m_{e}^{2}}$

$=\frac{\left(\frac{6.63 \times 10^{-34}}{2 \times 3.14}\right)^{2}}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times\left(9.1 \times 10^{-31}\right)^{2}}$$\approx 1.21 \times 10^{29} \,m$

It is known that the universe is $156$ billion light years wide or $1.5 \times 10^{27} \,m$ wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.