Bohr Model: Energy Levels and Spectra for Hydrogen-like Atoms

The Mathematical Engine: From Postulates to Predictions

The beauty of Bohr's model is that it provides just enough input to build a complete quantitative picture. We start with two key ideas: one from classical physics and one from Bohr's new quantum rules. For generality, we will consider a "hydrogen-like" atom: a nucleus with charge $+Ze$ and a single orbiting electron with charge $-e$. For hydrogen, $Z=1$.

1. Classical Force Balance: The electron is held in a circular orbit by the electrostatic Coulomb force, which provides the necessary centripetal force.
   $\frac{1}{4\pi {ϵ}_0}\frac{(Ze)e}{r^2}=\frac{m_e{v}^2}{r}$
   (For simplicity, we often use Coulomb's constant $k=1/(4\pi {ϵ}_0)$, so the equation becomes $\frac{kZ{e}^2}{r^2}=\frac{m_e{v}^2}{r}$).

2. Quantum Condition: The angular momentum of the electron is quantized.
   $L=m_{e}vr=n\frac{h}{2\pi }=n\hslash ,\text{where }n=1,2,3,\dots$

We now have a system of two equations with two unknowns: the orbital radius ($r$) and the electron's velocity ($v$). By solving this system, we can determine the allowed values for $r$ and, from there, the allowed energies.

The following video provides a clear, step-by-step walkthrough of this derivation. It will show you how to combine these two fundamental equations to find expressions for the quantized radius and energy of the electron.

To complement the video, here is a textual resource that lays out the same derivation, but includes the nuclear charge $Z$ from the start, making it applicable to all hydrogen-like atoms.

Let's summarize the two monumental results of this derivation:

1. Quantized Radii: The allowed orbital radii are not continuous. They are given by:
   $r_n=\frac{n^2}{Z}{a}_B$
   where $a_B=\frac{4\pi {ϵ}_0{\hslash }^2}{m_e{e}^2}\approx 0.0529\text{ nm}$ is the Bohr radius. This is the radius of the hydrogen atom ($n=1,Z=1$) in its lowest energy state.

2. Quantized Energy: The total energy of the electron in an allowed orbit is also quantized:
   $E_n=-\frac{Z^2}{n^2}{E}_0$
   where $E_0=\frac{m_e{e}^4}{2(4\pi {ϵ}_0{)}^2{\hslash }^2}\approx 13.6\text{ eV}$ is the ground-state energy of hydrogen.

Before we proceed, take a moment to reflect. In your own words, what is the physical meaning of the negative sign in the energy formula? What would a total energy of $E=0$ or $E>0$ signify for the electron?

The Moment of Triumph: Deriving the Rydberg Formula

Now we have the allowed energy levels. The final step is to connect this to the observed spectral lines using Bohr's third postulate: an electron emits a photon when it jumps from a higher energy state ($n_i$) to a lower one ($n_f$). The photon's energy must equal the energy difference:

$E_{\text{photon}}=\Delta E=E_i-E_f$

Since $E_{\text{photon}}=hf=hc/\lambda$, we can write:

$\frac{hc}{\lambda }=E_i-E_f=\left(-\frac{Z^2{E}_0}{n_i^2}\right)-\left(-\frac{Z^2{E}_0}{n_f^2}\right)$

Rearranging this gives:

$\frac{1}{\lambda }=\frac{Z^2{E}_0}{hc}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

This is precisely the form of the Rydberg equation! Bohr's model didn't just qualitatively explain line spectra; it derived the exact mathematical formula from first principles. The Rydberg constant, $R_H$, which was previously just an experimentally measured value, is now revealed to be a composite of more fundamental constants: $R_H=E_0/hc$. This was a staggering success.

The following video segment and image illustrate how these energy level transitions produce the famous spectral series of hydrogen.

Guided Discovery: Exploring the Model

Now, let's use these formulas to build your own intuition. Work through the following problems. The goal isn't just to get the right number, but to understand why the number makes sense.

Problem 1: The Balmer Series

The first line of the Balmer series for hydrogen ($Z=1$) corresponds to the transition from $n=3$ to $n=2$.

a) Before calculating, predict whether this photon will have more or less energy than the first line of the Lyman series ($n=2\to n=1$). Justify your prediction based on the energy level diagram.
b) Calculate the energy (in eV) and the wavelength (in nm) of the photon emitted in this $n=3\to n=2$ transition.
c) What region of the electromagnetic spectrum does this correspond to? Does this match the experimental observation of the Balmer series?

Insight:
The energy difference between levels shrinks as $n$ increases ($1/n^2$ dependence). The jump from $n=2\to n=1$ is the largest possible for a single-step de-excitation, so the Lyman series contains the highest energy photons. The Balmer series involves smaller energy drops, placing it at longer wavelengths. Your calculation for the $n=3\to n=2$ transition should yield a wavelength of about 656 nm, which is red light—the most prominent line in the visible hydrogen spectrum, exactly as observed.

Problem 2: Ionization from an Excited State

Consider a hydrogen atom where the electron is in the $n=4$ excited state.

a) How much energy, in eV, is required to ionize this atom?
b) Compare this value to the 13.6 eV required to ionize hydrogen from its ground state. Why is it so much easier to ionize an excited atom? Relate your answer to both the energy and the radius of the $n=4$ orbit.

Insight:
The energy of the $n=4$ state is $E_4=-13.6/4^2=-0.85$ eV. Thus, only 0.85 eV is needed to free the electron. The radius of this orbit is $r_4=4^2{a}_B=16a_B$. The electron is much farther from the nucleus and therefore much less tightly bound, requiring significantly less energy to be removed. This principle is why highly excited atoms ("Rydberg atoms") are very sensitive and easily ionized.

Problem 3: A Hydrogen-like Ion

Now consider a singly ionized Helium atom, ${\text{He}}^{+}$. This ion has a nucleus with $Z=2$ and a single electron.

a) What is the ground state ($n=1$) energy of ${\text{He}}^{+}$? How does this compare to hydrogen?
b) Calculate the wavelength of the photon emitted when the electron in ${\text{He}}^{+}$ transitions from $n=2$ to $n=1$.
c) In which region of the EM spectrum does this photon lie? Compare this to the equivalent transition in hydrogen (the first Lyman line, $\lambda \approx 122\text{ nm}$). What does the $Z^2$ dependence imply about the spectra of heavier ions?

Insight:
a) The energy is $E_1=-2^2/1^2\times 13.6\text{ eV}=-54.4\text{ eV}$. It is four times more strongly bound than hydrogen's electron.
b) The energy difference is $\Delta E=-54.4\left(\frac{1}{2^2}-\frac{1}{1^2}\right)=40.8\text{ eV}$. The wavelength is $\lambda =hc/\Delta E\approx 30.4\text{ nm}$.
c) This is in the extreme ultraviolet (EUV) region. The energy of transitions scales as $Z^2$, so the emitted photons from heavier hydrogen-like ions are much more energetic, shifting their spectra from the visible/UV into the EUV and X-ray regions. The stronger nuclear pull creates a deeper potential well with larger energy gaps between levels.

Conclusion

Today, we witnessed the quantitative power of Bohr's model. By taking his postulates as our engine, we successfully derived expressions for the discrete radii and energy levels of hydrogen-like atoms. Most importantly, we recovered the empirical Rydberg formula from first principles, providing a theoretical origin for the Rydberg constant.

Key Takeaways:

• The combination of classical force balance and quantized angular momentum yields quantized radii ($r_n\propto n^2/Z$) and energies ($E_n\propto -Z^2/n^2$).
• The negative energy of an electron indicates a bound state, with the magnitude representing the energy needed to ionize the atom from that state (the binding energy).
• The Bohr model correctly predicts the wavelengths of spectral lines in hydrogen-like atoms by equating the energy difference between stationary states to the energy of the emitted photon.
• The model's greatest triumph was deriving the Rydberg formula, explaining that the constant $R_H$ is determined by fundamental physical constants ($m_e,e,h,c$).

Despite this incredible success, the Bohr model is not the final word. It works almost perfectly for single-electron systems like H and ${\text{He}}^{+}$, but it fails dramatically when applied to atoms with multiple electrons. It also cannot explain why some spectral lines are brighter than others or what happens when atoms are placed in a magnetic field. In our next lesson, we will examine these failures, which reveal the conceptual cracks in Bohr's model and highlight why, despite its partial success, it is ultimately unsatisfying and must be replaced by a more complete theory.