First Bohr Radius - Quantum and Atomic Model

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SUMMARY

The discussion focuses on calculating the first Bohr radius of an electron bound to a proton by gravitational force instead of electric force. The relevant equation is modified from the electric force formula, resulting in r = Gmm/mv^2, where G represents the gravitational constant. Participants clarify that the electron orbits the proton, leading to the cancellation of the electron's mass in the equation. This manipulation allows for the determination of the radius of the first Bohr orbit under gravitational influence.

PREREQUISITES
  • Understanding of classical mechanics, specifically Newton's laws of motion.
  • Familiarity with gravitational force equations, particularly F = Gm1m2/r^2.
  • Knowledge of the original Bohr model of the hydrogen atom and its equations.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Explore the implications of gravitational binding in atomic models.
  • Research the differences between electric and gravitational forces in atomic structures.
  • Learn about the gravitational constant G and its significance in physics.
  • Investigate advanced topics in quantum mechanics related to atomic orbits.
USEFUL FOR

Students studying quantum mechanics, physics enthusiasts, and educators looking to deepen their understanding of atomic models and forces in nature.

PeachBanana
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Homework Statement



Suppose an electron was bound to a proton, as in the hydrogen atom, but by the gravitational force rather than by the electric force. What would be the radius of the first Bohr orbit?

Homework Equations



r = h^2 / 4∏^2*mke^2

The Attempt at a Solution



I'm not sure how to manipulate the above equation to account for gravitational force. Should I start by looking at that fact

F = ma
kZe^2/r^2 = mv^2/r
 
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Replace the electric force by the gravitational one. What is the force of gravity between two masses? (It is the mass of the proton and the mass of the electron now).

ehild
 
Ah, so F = km1m2/r^2 ?

Which means...

F = ma
km1m2/r^2 = mv^2/r
 
"k" must be the gravitational constant, usually denoted by G.
So what do you get for r?

ehild
 
r = Gmm/mv^2

The question I now have is which mass will cancel as I have a mass in the numerator and denominator...
 
Which particle orbits around the other?

ehild
 
The electrons orbit around the protons so does this mean the electron mass will cancel out?
 
PeachBanana said:
The electrons orbit around the protons so does this mean the electron mass will cancel out?

It needs the centripetal force, so the term mv^2/r refers to the
electron.


ehild
 

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