Speed of electron in Bohr model

  • Thread starter vaizard
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  • #1
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Homework Statement


In the classical Bohr model of the hydrogen atom, the nucleus has one positive electronic charge Q_n = e = +1.602E-19 Coulombs and the single electron (mass = 9.1E-31 kg and Q_e = -e = -1.602E-19 C, one negative electronic charge) orbits the nucleus in a circular orbit. In the Bohr model, stable orbits result only when the angular momentum is an integral multiple of h/(2\pi) (remember, L = r x p, but for circular orbits r is perpendicular to p, so here L = mvr = n(h/2\pi) and h = Planck's constant = 6.626E-34 J-s.

a) Calculate the speed with which the electron must orbit the proton for the ground state (n = 1) in the Bohr model of the hydrogen atom.

b) What is the radius of this ground state orbit? You probably know from Chemistry that the atomic size is about an Angstrom (10E-10 m). Is your answer close to this?

Homework Equations


[tex]L = mvr = \frac{nh}{2\pi}[/tex]
Not sure if this one is needed, but:
[tex]F = \frac{1}{4\pi\epsilon_0} \frac{|q_1q_2|}{r^2}[/tex]

The Attempt at a Solution


From the first equation,
[tex]v = \frac{nh}{2\pi mr}[/tex]
I know what n, h, and m are, but I don't have r (well, I know it's .529E-10, but I can't use it). For this reason, I'm trying to use the second equation to find r from the charge values for the proton and electron, but I don't know F, the force on each charge. Any ideas?

Thanks!
 

Answers and Replies

  • #2
Nabeshin
Science Advisor
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Remember from circular motion that [tex]a=\frac{v^{2}}{r}[/tex].

So Newton's 2nd law can be written as [tex]F=m\frac{v^{2}}{r}[/tex]. This should help :)
 
  • #3
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Yeah, I thought of that too, but once again, you need the radius to find the force by that method.
 
  • #4
378
2
two equations, two unknowns?

You have one equation, now use forces to get second.
will get something like
v^2 = k1/r
v = k2 /r
 
  • #5
43
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Well, you have Bohr'd second postulate as,
[tex]L = mvr = \frac{nh}{2\pi}[/tex]
On squaring it, you'll have:
[tex] m^2v^2r^2 = \frac{n^2h^2}{4\pi^2}[/tex]

Also, from the first postulate you know that,
Centripetal Force= Electrostatic Force
Therefore you'll have,
[tex]\frac{mv^2}{r}=~\frac{e^2}{4\pi\epsilon_o r^2} [/tex]

Now, equate the [tex]v^2[/tex] terms, with their appropriate values from the above two equations, you'll get an equation for r!

I hope this proves to be of help to you. :wink:
 
  • #6
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Thanks a lot, that worked perfectly! I would have never thought of that!

When I completed part B, I got an answer of [tex]\sim 0.48 \, \AA[/tex] for the radius, which is close enough to [tex]a_0 = 0.529 \, \AA[/tex]
 
  • #7
43
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Well, I'm always hereto help!
 

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