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Bohr's semiclassical aproach

by NEWO
Tags: aproach, bohr, semiclassical
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NEWO
#1
Jan21-06, 11:37 AM
P: 97
Hi there guys,

I just need something clarifying if possible.

Is the Bohr semiclassical aprroach to the energy levels of the hydrogen atom, Where in Schrodingers equation, the potential is the coulomb potential??? If you could clarify what the semiclassical approach it I would be very grateful

Thanks in advance

newo
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ZapperZ
#2
Jan21-06, 11:42 AM
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Quote Quote by NEWO
Hi there guys,
I just need something clarifying if possible.
Is the Bohr semiclassical aprroach to the energy levels of the hydrogen atom, Where in Schrodingers equation, the potential is the coulomb potential??? If you could clarify what the semiclassical approach it I would be very grateful
Thanks in advance
newo
Could you please reformulate your question in a clearer fashion? Are you asking about the Bohr model? Or are you asking about the Schrodinger equation? The Bohr model does not use the Schrodinger Equation.

Zz.
NEWO
#3
Jan21-06, 11:50 AM
P: 97
ah ok yes silly me

I will ask you in a more suitable question

I need to be able to derive the energy levels of the hydrogen atom by the use of Bohr semiclassical approach

and in order to do this my understanding of the semiclassical approach is very cloudy and the books I have read don't seem to make it clearer

I thank you for any help

Newo

Gokul43201
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Jan21-06, 11:59 AM
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Bohr's semiclassical aproach

What makes the Bohr model "semi"-classical is that he required that the angular momentum of the electron be quantized (integer multiples of [itex]\hbar[/itex]).

Using this requirement : [tex] L = mvr = n\hbar [/tex]

and the classical equation for circular motion under a central (electrostatic) force :

[tex]F_{centripetal} = mv^2/r = ke^2/r^2 [/tex]

gives you Bohr's results for the energies, radii and orbital velocities of the different orbits.
NEWO
#5
Jan21-06, 12:02 PM
P: 97
ok thankyou for that Goku143201,
how can this be used for the derivation of the energy levels of the hydrogen atom??
thanks
NEWO
#6
Jan21-06, 12:54 PM
P: 97
its okay now I found out how to do it you get the answer as being

[tex] [13.5/n^2][/tex]


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