# How to find the lifetime of sodium atoms in the 3p using information on the spectral line width?

I am trying to find the lifetime of sodium atoms at the 3p state using the relationship between the spectral linewidth and the uncertainty principle: $$\Delta E \times \Delta t = h / 4\pi$$ The information I have is the full-width-half-maximum (FWHM) linewidth of the sodium atomic absorption spectrum $$\lambda = \pu{9.2E-14 m}$$. How would I use this equation to find the lifetime?

The atomic transition from ground 3s to excited 3p states has a wavelength $$\lambda = \pu{589.0 nm}$$.

I used the relationship $$\Delta E = \frac{h c \Delta \lambda}{\lambda^{2}}$$

Plugging in $$\Delta \lambda = \pu{9.200E-14 m}$$, then $$\Delta E = \frac{h \cdot \pu{3E8 m} \cdot \pu{9.200E-14 m}} {(\pu{589.0E-9 m})^{2}}$$

So using the uncertainty relationship, we have that $$\Delta t = \frac{h}{ 4 \pi \Delta E} = \frac{(\pu{589.0E-9 m})^{2}}{4 \pi (\pu{3E8 m s^-1} \cdot \pu{9.200E-14 m})} = \pu{1.000E9 s}$$

My attempt to convert the linewidth from unit of $$\pu{nm}$$ to $$\pu{eV}$$: $$E = \frac{hc}{\lambda} = \frac{\pu{6.626E-34 J s} \cdot \pu{3E8 ms^-1}}{\pu{9.200E-14 m}} = \pu{2.161E-12 J} = \pu{1.348E7 eV}$$

• I find something in 10^-47 s. Am I wrong ? Does it make sense ? – Maurice Jan 22 at 11:18
• @Maurice It is to be expected that the number will be much smaller than the literature value (17 ns), because the linewidth is smaller in the problem. However, I calculted $1 \times 10^{-9}$ s. Would you mind sharing how you calculated the time? Thank you. – ferris Jan 22 at 11:48
• The answer suggested in the comment as 10^-47 s is completely wrong. Your answer is in the right ball-park of ns. Make sure you don't round off anywhere until the end. hyperphysics.phy-astr.gsu.edu/hbase/quantum/parlif.html – M. Farooq Jan 22 at 15:09
• @M.Farooq How do you convert the linewidth from nm to eV? I tried $E = hc/ \lambda$ and changed the unit from J to eV, but the result is orders of magnitude away from the number on that website. – ferris Jan 22 at 19:46
• Your units must be inconsistent. If you are using $\delta$E in eV your Planck's constant must be in eV as well. Please show your calculations and steps in the original post and we can start from there. – M. Farooq Jan 22 at 19:49

Half the fwhm has to be added/subtracted to the central wavelength and then this change in wavelength converted to an energy. Thus the limits on the wavelength are $$\lambda \pm \Delta \lambda /2)$$ and the energy change $$\Delta E=hc/(\lambda - \Delta \lambda /2)-hc/(\lambda + \Delta \lambda /2)$$ Joules with $$c$$ in m/s and $$h$$ in J s. Plugging in the numbers gives a lifetime of $$\approx 10^{-9}$$, i.e. $$1$$ ns.
As a rule of thumb, lifetimes of excited electronic states are rarely less than $$10^{-12}$$s and if the transition is allowed, as in this case often of the order of nanoseconds to microseconds. If transitions are forbidden then the lifetimes can extend from microseconds to seconds.
• Interesting that I missed $\Delta \lambda /2$. However, I don't recall seeing this as such. If $\Delta$E is uncertainity, what is the rigorous statistical basis of dividing FWHM by 2? It is not equal to the standard deviation for a Gaussian, Voigt or Lorentz profile. – M. Farooq Jan 24 at 16:27
• FWHM is full width at half maximum so divide by 2 and then work out energy spread. The important thing is that $\Delta \lambda$ does not directly give the change in energy. FWHM is not 1/e of width so this will not give absolutely correct answer, however using width to get lifetimes is only really true if nothing interacts with system so it is more hypothetical than real. Anyway the OP should only use the data in the question. – porphyrin Jan 25 at 9:17