First Bohr Radius - Quantum and Atomic Model

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Homework Help Overview

The problem involves determining the radius of the first Bohr orbit for an electron bound to a proton by gravitational force instead of electric force, as in a hydrogen atom. The context is rooted in quantum mechanics and atomic models.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss manipulating the equation for the radius of the Bohr orbit to account for gravitational force, questioning how to replace electric force with gravitational force. There are attempts to clarify the relationship between the forces acting on the electron and proton.

Discussion Status

Participants are actively engaging with the problem, exploring the implications of using gravitational force. There is a focus on the roles of the masses involved and how they affect the calculations. Some guidance has been offered regarding the gravitational constant and the forces at play, but no consensus has been reached on the final expression for the radius.

Contextual Notes

Participants are considering the implications of using gravitational force in place of electric force, which may involve assumptions about the behavior of the masses in the system. The discussion reflects on the need to clarify which mass will cancel in the equations being used.

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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