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fluidistic
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Homework Statement
1)Determine the radius of the allowed orbits. Calculate the first orbit of Bohr's model for the hydrogen atom.
2)Show that the energy is quantized. Calculate the energy of an electron on the first orbit (fundamental state of hydrogen atom)
Homework Equations
[itex]L=n \hbar[/itex].
[itex]F=\frac{k(e^-) ^2}{R^2}[/itex]
[itex]F=m_e v^2/R[/itex]
[itex]k=\frac{1}{4\pi \varepsilon _0}[/itex].
The Attempt at a Solution
I used ideas of classical mechanics, combined with the supposition that the electron orbiting the nucleus doesn't emit photons and that the angular momentum is quantized.
I found out that [itex]v=\frac{n \hbar}{m_e R}[/itex] and that [itex]R=\frac{4\pi \varepsilon _0 n^2 \hbar ^2}{(e^-)^2m_e}[/itex]. I checked out in wikipedia and taking n=1 I indeed find the Bohr's radius, that's why I don't show all my arithmetics here.
I get that for n=1, [itex]R\approx 5.29177208 \times 10 ^{-11}m[/itex].
And that the velocity of the electron is about [itex]2187691.254 \frac{m}{s}[/itex], which is less than 10% of light's speed. Thus I can consider the classical momentum [itex]p=m_ev[/itex] as a good approximation for the momentum value of the electron. Or I can take [itex]\frac{m_e v^2}{2}[/itex] for the kinetic energy of the electron.
Precisely, this is taking this expression for the kinetic energy of the electron that I'm getting a problem for part 2).
So I answered part 1).
Now for part 2, E=T+V, where [itex]V=-\frac{k(e^-)^2}{R}[/itex].
Replacing T by [itex]\frac{m_e v^2}{2}[/itex], I reach that [itex]E=\frac{(e^-)^4m_e}{16 \pi ^2 \varepsilon _0 ^2n^2\hbar ^2}-\frac{(e^-)^4m_e}{16 \pi ^2 \varepsilon _0 ^2n^2\hbar ^2}=0[/itex].
In other words, the kinetic energy of the electron is worth minus the potential energy of the electron and thus its total energy is null. I know this is a wrong answer, I should find -13.6 eV for n=1, but I don't reach this.
Any help to spot my error(s) is welcome. I'm stuck on this since yesterday.
Edit: Nevermind, I found an algebra mistake for T by a factor 1/2. Now I think I'll reach the results.
Edit 2: Indeed, I find something (my calculator can't do the math, it shows E=0 because I'm guessing of an [itex]10^{-106}[/itex]) by hand that is of the order of [itex]-10^{-18}J[/itex] which is roughly of the order of -10 eV (compared to -13.6 eV) so I'm guessing that my expression is right.
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