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

Question

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 $\pu{eV}$: $$ E = \frac{hc}{\lambda} = \frac{\pu{6.626E-34 J s} \cdot \pu{3E8 ms^-1}}{\pu{9.200E-14 m}} = \pu{2.161E-12 J} = \pu{1.348E7 eV} $$

Answer 1

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.