Bohr’s analysis of the hydrogen atom seemed to be a resounding success. By introducing Einstein’s ideas about light quanta, Bohr was able to provide the first understanding of discrete spectra and to predict the Balmer formula for the wavelengths in the hydrogen spectrum. And the Bohr hydrogen atom, unlike Rutherford’s model, was stable. There was clearly some validity to the idea of stationary states.

But Bohr was completely unsuccessful at explaining the spectra of any other neutral atom. His method did not work even for helium, the second element in the periodic table with a mere two electrons. Something inherent in Bohr’s assumptions seemed to work correctly for a single electron but not in situations with two or more electrons.

It is important to make a distinction between the Bohr model of atomic quantization, described in Section 38.5, and the Bohr hydrogen atom. The Bohr model assumes that stationary states exist, but it does not say how to find them. We found the stationary states of a hydrogen atom by requiring that an integer number of de Broglie waves fit around the circumference of the orbit, setting up standing waves. The difficulty with more complex atoms is not the Bohr model but the method of finding the stationary states. Bohr’s model of the atomic quantization remains valid, and we will continue to use it, but the procedure of fitting standing waves to a circle is just too simple to find the stationary states of complex atoms. We need to find a better procedure.

Einstein, de Broglie, and Bohr carried physics into uncharted waters. Their successes made it clear that the microscopic realm of light and atoms is governed by quantization, discreteness, and a blurring of the distinction between particles and waves. Although Bohr was clearly on the right track, his inability to extend the Bohr hydrogen atom to more complex atoms made it equally clear that the complete and correct theory remained to be discovered. Bohr’s theory was what we now call “semiclassical,” a hybrid of classical Newtonian mechanics with the new ideas of quanta. Still missing was a complete theory of motion and dynamics in a quantized universe—a quantum mechanics.

CHAPTER 38 CHALLENGE EXAMPLE Hydrogen fluorescence

Fluorescence is the absorption of light at one wavelength followed by emission at a longer wavelength. Suppose a hydrogen atom in its ground state absorbs an ultraviolet photon with a wavelength of 95.10 nm. Immediately after the absorption, the atom undergoes a transition with $\text{Δ}n=3$. What is the wavelength of the photon emitted in this transition?

MODEL Photons are emitted and absorbed as an atom undergoes transitions from one energy level to another. The Bohr model gives the energy levels of the hydrogen atom.

VISUALIZE FIGURE 38.23 shows the process. To be absorbed, the photon energy has to match exactly the energy difference between the ground state of hydrogen and an excited state with quantum number $n$. After excitation, the atom emits a photon as it jumps downward in a $n\to n-3$ transition.

FIGURE 38.23 The process of fluorescence in hydrogen. Energy levels are not drawn to scale.

FIGURE 38.23 Full Alternative Text

SOLVE The energy of the absorbed photon is

$$E=hf={\displaystyle \frac{hc}{\lambda }}={\displaystyle \frac{(4.14\times {10}^{-15}\text{eV}\text{s})(3.00\times {10}^8\text{m}\text{/}\text{s})}{95.10\times {10}^{-9}\text{m}}}=13.06\text{eV}$$

The atom’s initial energy is $E_1=-13.60\text{eV,}$ the energy of the ground state of hydrogen. Absorbing a 13.06 eV photon raises the atom’s energy to $E_n=E_1+13.06\text{eV}=-0.54\text{eV}$. The energy levels of hydrogen are given by

$$E_n=-{\displaystyle \frac{13.60\text{eV}}{n^2}}$$

The quantum number of the energy level with $-0.54\text{eV}$ is

$$n=\sqrt{-{\displaystyle \frac{13.60\text{eV}}{(-0.54\text{eV})}}}=5$$

We see that the absorption is a $1\to 5$ transition; thus the emission, with $\text{Δ}n=3$, must be a $5\to 2$ transition. The energy of the $n=2$ state is

$$E_2=-{\displaystyle \frac{13.60\text{eV}}{2^2}}=-3.40\text{eV}$$

Consequently, the energy of the emitted photon is

$$E_{\text{photon}}=\text{Δ}{E}_{\text{atom}}=(-0.54\text{eV})-(-3.40\text{eV})=2.86\text{eV}$$

Inverting the energy-wavelength relationship that we started with, we find

$$\lambda ={\displaystyle \frac{hc}{E_{\text{photon}}}}={\displaystyle \frac{(4.14\times {10}^{-15}\text{eV}\text{s})(3.00\times {10}^8\text{m}\text{/}\text{s})}{2.86\text{eV}}}=434\text{nm}$$

When atomic hydrogen gas is irradiated with ultraviolet light having a wavelength of 95.10 nm, it fluoresces at the visible wavelength of 434 nm. (It also fluoresces at infrared and ultraviolet wavelengths in downward transitions with other values of $\text{Δ}n\text{.})$

REVIEW The $5\to 2$ transition is a member of the Balmer series, and a 434 nm spectral line was shown in the hydrogen spectrum of Figure 37.7. It is important to notice that the 13.06 eV photon energy does not match any energy level of the hydrogen atom. Instead, it matches the difference between two levels because that conserves energy in a transition between those two levels. Photons with nearby wavelengths, such as 94 nm or 96 nm, would not be absorbed at all because their energy does not match the difference of any two energy levels in hydrogen.