COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 26, Problem 94QAP
To determine

(a)

The radius of the first Bohr orbit of the hydrogen atom considering that the electron is bound to the proton by gravitational force (rather than electrostatic force)

Expert Solution
Check Mark

Answer to Problem 94QAP

The radius of the first Bohr orbit of the hydrogen atom is given by the following expression r=h24π2Gme2mp

Explanation of Solution

Given:

It is assumed that the electron is bound to the proton by gravitational force (rather than electrostatic force) in hydrogen atom.

Formula used:

In this case the centripetal force on the electron is equal to the gravitational force of attraction between the proton and electron. Therefore we can write

  mev2r=Gmempr2

where me and mp are electron and proton mass respectively, G is the gravitational constant.

According to Bohr quantization principle angular momentum of an electron in an atom is an integral multiple of h/2p.

So mathematically we can write it as

  L=mevr=nh2π

where, me is the mass of the electron, v and r are the velocity of the electron and radius of the electron orbit respectively, n is a positive integer.

Calculation:

From Eq. (1.1) we can write

  v2r=Gmp

Putting n =1 for the 1st Bohr orbit in Eq. (1.2) we get

  mevr=h2πor, v =h2πmer

Using the expression for v in Eq. (1.3) we get

  h24π2me2r2r=Gmpor, r =h24π2Gme2mp

Conclusion:

Therefore, the radius of the first Bohr orbit of the hydrogen atom is given by the following expression r =h24π2Gme2mp

To determine

(b)

The energy of the electron in the first Bohr orbit

Expert Solution
Check Mark

Answer to Problem 94QAP

The energy expression for the electron in the 1st Bohr orbit is given by E=2π2G2me3mp2h2

Explanation of Solution

Given:

It is assumed that the electron is bound to the proton by gravitational force (rather than electrostatic force) in hydrogen atom.

Calculation:

Now substituting the expression for the radius of the 1st Bohr orbit r given by Eq. 1.5 in Eq. 1.4 we get the expression for the velocity of the electron as

  v=2πGmemph

So the kinetic energy of the electron is

  Ek=12me( 2πG m e m p h)2=2π2G2me3mp2h2

The gravitational potential energy of the electron is given by

  Ep=Gmempr

Now substituting the expression for r from Eq. 1.5 in Eq. 1.8 we get

  Ep=4π2G2me3mp2h2

Therefore the total energy of the electron in the 1st Bohr orbit is given by

  E=Ek+Ep

Now using the expressions for Ek (Eq. 1.7) and Ep (Eq. 1.9) we get the total energy as

  E=2π2G2me3mp2h2

Conclusion:

Therefore, the energy expression for the electron in the 1st Bohr orbit is given by E=2π2G2me3mp2h2

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Chapter 26 Solutions

COLLEGE PHYSICS

Ch. 26 - Prob. 11QAPCh. 26 - Prob. 12QAPCh. 26 - Prob. 13QAPCh. 26 - Prob. 14QAPCh. 26 - Prob. 15QAPCh. 26 - Prob. 16QAPCh. 26 - Prob. 17QAPCh. 26 - Prob. 18QAPCh. 26 - Prob. 19QAPCh. 26 - Prob. 20QAPCh. 26 - Prob. 21QAPCh. 26 - Prob. 22QAPCh. 26 - Prob. 23QAPCh. 26 - Prob. 24QAPCh. 26 - Prob. 25QAPCh. 26 - Prob. 26QAPCh. 26 - Prob. 27QAPCh. 26 - Prob. 28QAPCh. 26 - Prob. 29QAPCh. 26 - Prob. 30QAPCh. 26 - Prob. 31QAPCh. 26 - Prob. 32QAPCh. 26 - Prob. 33QAPCh. 26 - Prob. 34QAPCh. 26 - Prob. 35QAPCh. 26 - Prob. 36QAPCh. 26 - Prob. 37QAPCh. 26 - Prob. 38QAPCh. 26 - Prob. 39QAPCh. 26 - Prob. 40QAPCh. 26 - Prob. 41QAPCh. 26 - Prob. 42QAPCh. 26 - Prob. 43QAPCh. 26 - Prob. 44QAPCh. 26 - Prob. 45QAPCh. 26 - Prob. 46QAPCh. 26 - Prob. 47QAPCh. 26 - Prob. 48QAPCh. 26 - Prob. 49QAPCh. 26 - Prob. 50QAPCh. 26 - Prob. 51QAPCh. 26 - Prob. 52QAPCh. 26 - Prob. 53QAPCh. 26 - Prob. 54QAPCh. 26 - Prob. 55QAPCh. 26 - Prob. 56QAPCh. 26 - Prob. 57QAPCh. 26 - Prob. 58QAPCh. 26 - Prob. 59QAPCh. 26 - Prob. 60QAPCh. 26 - Prob. 61QAPCh. 26 - Prob. 62QAPCh. 26 - Prob. 63QAPCh. 26 - Prob. 64QAPCh. 26 - Prob. 65QAPCh. 26 - Prob. 66QAPCh. 26 - Prob. 67QAPCh. 26 - Prob. 68QAPCh. 26 - Prob. 69QAPCh. 26 - Prob. 70QAPCh. 26 - Prob. 71QAPCh. 26 - Prob. 72QAPCh. 26 - Prob. 73QAPCh. 26 - Prob. 74QAPCh. 26 - Prob. 75QAPCh. 26 - Prob. 76QAPCh. 26 - Prob. 77QAPCh. 26 - Prob. 78QAPCh. 26 - Prob. 79QAPCh. 26 - Prob. 80QAPCh. 26 - Prob. 81QAPCh. 26 - Prob. 82QAPCh. 26 - Prob. 83QAPCh. 26 - Prob. 84QAPCh. 26 - Prob. 85QAPCh. 26 - Prob. 86QAPCh. 26 - Prob. 87QAPCh. 26 - Prob. 88QAPCh. 26 - Prob. 89QAPCh. 26 - Prob. 90QAPCh. 26 - Prob. 91QAPCh. 26 - Prob. 92QAPCh. 26 - Prob. 93QAPCh. 26 - Prob. 94QAPCh. 26 - Prob. 95QAPCh. 26 - Prob. 96QAPCh. 26 - Prob. 97QAPCh. 26 - Prob. 98QAPCh. 26 - Prob. 99QAPCh. 26 - Prob. 100QAPCh. 26 - Prob. 101QAPCh. 26 - Prob. 102QAPCh. 26 - Prob. 103QAPCh. 26 - Prob. 104QAP
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