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On the Wikipedia I found that the Bohr radius is equal to:

\begin{align} \boxed{r_b=\dfrac{4\pi\varepsilon_0\hbar^2}{m_e{e}^2}} \end{align}

but while we have been learning Bohr's model we derived the equation for a radius of the electron orbit like they do it on the hyperphysics:

\begin{split} &2\pi r = N \lambda_b\\ &\phantom{2\pi}r = \frac{N \lambda_b}{2\pi}\longleftarrow \lambda_b = \tfrac{h}{p}~,~p=mv\gamma(v)\\ &\phantom{2\pi}r = \frac{N h}{2\pi m_e v\, \gamma(v)}\\ &\phantom{2\pi}r = \frac{Nh \sqrt{4 \pi \varepsilon_0 r m_e}}{2 \pi m_e Ze_0\, \gamma(v)}\longleftarrow v = \sqrt{\tfrac{Z{e_0}^2}{4\pi \varepsilon_0rm_e}}\\ &\phantom{\,~\,}r^2 = \frac{N^2h^2 4 \pi \varepsilon_0 r m_e}{4 \pi^2 {m_e}^2 Z{e_0}^2 {\gamma(v)}^2}\\ &\phantom{2\pi}r = \frac{N^2h^2 \varepsilon_0}{\pi m_e Z{e_0}^2 {\gamma(v)}^2}\\ \\ &\substack{\text{We assume that $v \ll c$ and therefore $\gamma(v) = 1$}}\\ \\ &~\,~\boxed{r = \dfrac{N^2h^2 \varepsilon_0}{\pi m_e Z{e_0}^2}}~\boxed{r=\dfrac{N^2}{Z\vphantom{{e_0}^2}}r_b}\xrightarrow{\text{this means that Bohr radius is} } \boxed{r_b = \dfrac{h^2\varepsilon_0}{\pi m_e {e_0}^2}} \end{split}

Ok so I have two equations for Bohr's radius but I can't connect them, because if I insert $\hbar = h/(2\pi)$ into the one from Wikipedia I don't get the one from hyperphysics:

\begin{align} r_b=\dfrac{4\pi\varepsilon_0\hbar^2}{m_e{e}^2} = \dfrac{4\pi\varepsilon_0h^2}{4^2\pi^2m_e{e}^2} = \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\dfrac{h^2\varepsilon_0}{4\pi m_e{e}^2}}_{\substack{\text{This is not the same as the}\\\text{equation on the Hyperphysics...}}} \end{align}

Is Wikipedia wrong? Please someone fix it if it is...

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2 Answers 2

up vote 2 down vote accepted

You made a mistake. On your last derivation, you have an extra 4 at the denominator:

\begin{align} r_b=\dfrac{4\pi\varepsilon_0\hbar^2}{m_e{e}^2} = \dfrac{4\pi\varepsilon_0h^2}{4\pi^2m_e{e}^2} = \frac{h^2\varepsilon_0}{\pi m_e{e}^2} \end{align}

Which is the formula you derived from hyperphysics.

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How can i be so sloppy $2^2 = 4!$ Sorry... –  71GA Jul 30 '13 at 10:20

So, essentially, Wikipedia has $$ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_{\mathrm{e}} e^2} \, , $$ and hyperphysics has $$ r = \frac{n^2 h^2 \varepsilon_0}{Z \pi m_{\mathrm{e}} e^2} = \frac{n^2 a_0}{Z} \, . $$

From the hyperphysics formula $$ a_0 = \frac{h^2 \varepsilon_0}{\pi m_{\mathrm{e}} e^2} \, , $$ substituting $$ h = 2 \pi \hbar \, , $$ one will get exactly what is written in Wikipedia $$ a_0 = \frac{4 \pi^2 \hbar^2 \varepsilon_0}{\pi m_{\mathrm{e}} e^2} = \frac{4 \pi \hbar^2 \varepsilon_0}{m_{\mathrm{e}} e^2} \, . $$

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