Question 40.7: The Line Width of Atomic Emissions Atoms have quantized ener......
The Line Width of Atomic Emissions
Atoms have quantized energy levels similar to those of Planck’s oscillators, although the energy levels of an atom are usually not evenly spaced. When an atom makes a transition between states, energy is emitted in the form of a photon. Although an excited atom can radiate at any time from t=0 to t=\infty, the average time interval after excitation during which an atom radiates is called the lifetime \tau. If \tau=1.0 \times 10^{-8} \mathrm{~s}, use the uncertainty principle to compute the line width \Delta f produced by this finite lifetime.
Conceptualize The lifetime \tau given for the excited state can be interpreted as the uncertainty \Delta t in the time at which the transition occurs. This uncertainty corresponds to a minimum uncertainty in the frequency of the radiated photon through the uncertainty principle.
Categorize We evaluate the result using concepts developed in this section, so we categorize this example as a substitution problem.
Use Equation 40.5 to relate the uncertainty in the photon’s frequency to the uncertainty in its energy:
E=h f \rightarrow \Delta E=h \Delta f \rightarrow \Delta f=\frac{\Delta E}{h}
Use Equation 40.24
\Delta E \Delta t \geq \frac{\hbar}{2 } (40.24)
to substitute for the uncertainty in the photon’s energy, giving the minimum value of \Delta f :
\Delta f \geq \frac{1}{h} \frac{\hbar}{2 \Delta t}=\frac{1}{h} \frac{h / 2 \pi}{2 \Delta t}=\frac{1}{4 \pi \Delta t}=\frac{1}{4 \pi \tau}
Substitute for the lifetime of the excited state:
\Delta f \geq \frac{1}{4 \pi\left(1.0 \times 10^{-8} \mathrm{~s}\right)}=8.0 \times 10^{6} \mathrm{~Hz}