# Doubt with another form of Balmer' Series

In the book I'm studying from (Solid state electronic devices - Ben Streetman) it is said that the Balmer series is defined as follows: $$v = cR \left( \frac{1}{2^2} - \frac{1}{n^2}\right)$$

with c speed of light and $R = 109,678 [cm]^{-1}$. But in the majority of other books and sites we can find the series as $$v = R_H \left( \frac{1}{2^2} - \frac{1}{n^2}\right)$$

with $R_H = 10973731.57 m^{−1}$ or, in its wavelength form: $$\lambda = b\left ( \frac{n^2}{n^2-4} \right)$$ with $b = 364.56 [nm]$

So, the problem is that I can't figure out a direct relation between the equation from Streetman's book and the latter. Basically if we work it out we should get that $\frac{4}{cR} = b$ or $\frac{1}{cR} = R_H$ but using $c = 2.998*10^{10} [cm/s]$ and $R = 109,678 (cm)^{-1}$ clearly what I said doesn't hold.

• The greek nu can both denote frequency or wavenumber. – Pieter Aug 26 '18 at 21:16
• May be set c=1 ? – R.G.J Aug 26 '18 at 22:23
• To be honest, not sure. In the book, they don't explicitly say c speed of light, it is assumed from the context and because before, in another equation, they said that c is the speed of light. So, if we assume here c = 1, not sure what would it mean in terms of physics. So far the only thing I could say is that the equation is simply wrong. I tried to find in other books and in the web a lot about this series, but none of them showed the above form. – Miguel Duran Diaz Aug 26 '18 at 22:27

## 1 Answer

The answer depends a little on history.

Firstly, the Bohr's model of the atom can be used to derive the theoretical prediction of the hydrogen atom transition wavelenghts as: $$\frac{1}{\lambda} = R\left(\frac{1}{n_L^2} - \frac{1}{n_U^2}\right)$$ In this case, $\lambda$ is the wavelength, and $R$ is the Rydberg constant defined as: $$R=\frac{me^4}{8 \epsilon_0^2 h^3 c}$$ If you substitute the numerical values for $m, c, e, h$ and $\epsilon_0$, you will get a value of $R \approx 10973731.5m^{-1}$. Note $c$ is the speed of light and $h$ is Planck constant.

Now, in Johann Balmer's time, the Bohr's model of the hydrogen atom was not available. Balmer's equation was $$\frac{1}{\lambda} = R_B \left(\frac{1}{2^2} - \frac{1}{n^2}\right)$$ where $R_B$ (due to Balmer) was experimentally determined (in order to fit the spectral observations) as: $R_B=109678cm^{-1}=10967800m^{-1}$. Note that $\frac{R}{R_B} \approx 1.00054$.

Now, notice that if we denote $\nu$ as the frequency in Hertz, then: $$\frac{1}{\lambda} = \frac{\nu}{c}= R \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \Rightarrow \nu = cR \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \approx cR_B \left(\frac{1}{2^2} - \frac{1}{n^2}\right)$$ where $c$ is the speed of light.

Summary: In your first equation $\nu$ refers to the frequency of the atomic transition. In the second equation $\nu$ is defined as $\nu=\frac{1}{\lambda}$. Also note the difference between the reciprocal of the wavelength, $\frac{1}{\lambda}$, and the wavenumber, $|\mathbf{k}|$, which is typically defined as $\frac{2\pi}{\lambda}$.