Class 12 Physics

CBSE and UP Board

Atoms

Chapter-12 Exercise 12.12

NCERT Solutions for Class 12 Physics Chapter 12 Question-12

# The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10⁻⁴⁰ .An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

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

r1 = 4πε0 (h/2π)²/m

_{e}e² ————- Eq-1Where,

Eo = Permittivity of free space

h = Planck’s constant = 6.63 x 10

^{-34}Jsm

_{e}= Mass of an electron = 9.1 x 10^{-³¹}kge = Charge of an electron = 1.9 x 10

^{–}^{19}Cm

_{p}= Mass of a proton = 1.67 x 10^{–}^{27}kgr= Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as:

F

_{C}= e²/4πε0r²————-Eq-2Gravitational force of attraction between an electron and a proton is given as:

F

_{G}=G m_{p}m_{e}/r² ——————- Eq-3 ,^{F}c, =-^H- -(3)Where, G = Gravitational constant = 6.67 x 10⁻¹¹ N m

^{2}/kg^{2}If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

Therefore , F

_{G}=F_{C}=> G m

_{p}m_{e}/r² = e²/4πε0r²Therefore ,

e²/4πε0 =G m

_{p}m_{e}—————Eq-4Putting the value of equation (4) in equation (1),we get:

r1= (h/2π)²/G m

_{p}m²_{e}= [6.63 x 10

^{-34}/(2 x 3.14)]²/(6.67 x 10⁻¹¹) x (1.67 x 10^{–}^{27}) x (9.1 x 10^{-³¹})²≈1.21 x 10²⁹

It is known that the universe is 156 billion light years wide or 1.5 x 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.