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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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Class 12 Physics
CBSE and UP Board
Atoms
Chapter-12 Exercise 12.12
NCERT Solutions for Class 12 Physics Chapter 12 Question-12

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1 Answer

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

    r1 = 4πε0  (h/2π)²/mee² ————- Eq-1

    Where,

    Eo = Permittivity of free space

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

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

    e = Charge of an electron = 1.9 x 1019 C

    mp= Mass of a proton = 1.67 x 10 27 kg

    r= Distance between the electron and the proton

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

    FC = e²/4πε0r²————-Eq-2

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

    FG =G mp me /r²   ——————- Eq-3                     ,

    Fc, =-^H-                                         -(3)

    Where, G = Gravitational constant = 6.67 x 10⁻¹¹ N m2/kg2

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

    Therefore , FG=FC

    =>    G mp me /r² = e²/4πε0r²

    Therefore ,

    e²/4πε0 =G mp me—————Eq-4

    Putting the value of equation (4) in equation (1),we get:

    r1= (h/2π)²/G mpe

    = [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

    ≈1.21 x 10²⁹
    It is known that the universe is 156 billion light years wide or 1.5 x 1027 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.

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