# Homework Help: Bohr's atom

1. Sep 30, 2011

### fluidistic

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

$L=n \hbar$.
$F=\frac{k(e^-) ^2}{R^2}$
$F=m_e v^2/R$
$k=\frac{1}{4\pi \varepsilon _0}$.
3. 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 $v=\frac{n \hbar}{m_e R}$ and that $R=\frac{4\pi \varepsilon _0 n^2 \hbar ^2}{(e^-)^2m_e}$. 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, $R\approx 5.29177208 \times 10 ^{-11}m$.
And that the velocity of the electron is about $2187691.254 \frac{m}{s}$, which is less than 10% of light's speed. Thus I can consider the classical momentum $p=m_ev$ as a good approximation for the momentum value of the electron. Or I can take $\frac{m_e v^2}{2}$ 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).
Now for part 2, E=T+V, where $V=-\frac{k(e^-)^2}{R}$.
Replacing T by $\frac{m_e v^2}{2}$, I reach that $E=\frac{(e^-)^4m_e}{16 \pi ^2 \varepsilon _0 ^2n^2\hbar ^2}-\frac{(e^-)^4m_e}{16 \pi ^2 \varepsilon _0 ^2n^2\hbar ^2}=0$.
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 $10^{-106}$) by hand that is of the order of $-10^{-18}J$ which is roughly of the order of -10 eV (compared to -13.6 eV) so I'm guessing that my expression is right.

Last edited: Sep 30, 2011
2. Sep 30, 2011

### DiracRules

It's correct, though you wrote it a bit confusingly :D

Since Coulomb force is the centripetal one,
$\frac{Ze^2}{4\pi\epsilon_0 r^2}=\frac{mv^2}{r}\Rightarrow mv^2=\frac{Ze}{4\pi\epsilon_0 r}$
Now, $E=K+V=\frac{mv^2}{2}-\frac{Ze}{4\pi\epsilon_0 r}=\frac{Ze}{2\cdot 4\pi\epsilon_0 r}-\frac{Ze}{4\pi\epsilon_0 r}=-\frac{Ze}{16\pi\epsilon_0 r}$

Substituting your expression for r in the previous equation, you get
$E_n=-\frac{Ze^2 }{16\pi\epsilon_0 \frac{4\pi\epsilon_0 n^2\hbar^2}{Ze^2 m}}=-Z^2 m\left(\frac{e^2}{8\pi\epsilon_0\hbar}\right)^2 \frac{1}{n^2}=-Z^2 \frac{R_H}{n^2}$ where $R_H$ is the Rydberg constant: $R_H=13.6eV$

3. Sep 30, 2011

### fluidistic

Thank you Mr. Dirac (may I ask you an autograph by the way?).
I am a bit confused with the Rydberg's constant. In wikipedia I read that $R_{\infty }$ in the case of the simplified hydrogen atom is worth $\frac{m_e (e^-)^4}{8 \varepsilon _0 ^2 h^3 c}$ which differs from our case.

4. Oct 1, 2011

### DiracRules

Probably my fault.

Here is the thing: the problem is that physicists usually tend to express all their constant in terms of energy (despite the original values are temperatures, wavelengths ...)
So, since it is easier to remember, I learnt that $R_\infty=13.6eV$, that is, the lowest energy level of the hydrogen atom.

NB: in the previous note, I didn't write correctly, since I should have written $R_\infty$ instead of $R_H$

Is that ok?