Natural broadening of lithium sub-levels

[gboff21]

Homework Statement

In the 2p state of lithium, two fine structure components are formed due to spin-orbit interaction. They produce wavelengths of 670.960nm and 670.975nm. Compare Doppler and natural broadening at 700K. The lifetime of the 2p state is 27ns and [itex]\frac{\Delta \lambda}{\lambda} ~= \frac{\Delta \nu}{\nu} [/itex] at [itex]\frac{\Delta \nu}{\nu} << 1[/itex]

Homework Equations

Given Doppler broadening: [itex]\Delta \nu = \frac{2 \nu_{0}}{c}\sqrt{\frac{2KT}{M}*ln(2)}[/itex]

The Attempt at a Solution

My Natural broadening derivation:
uncertainty principle: [itex]\Delta E \tau = \hbar[/itex] (tau is lifetime)
so [itex]\Delta \lambda = 2 \pi c \tau[/itex] (which comes out as ~50m)
This is clearly wrong since the doppler broadening comes out as in the order of 10^-12.

Any clues?

 

[mfb]

How did you get ##\Delta \lambda## based on ##\Delta E##? That relation depends on the wavelength itself.
It is easier to convert ##\Delta E## to ##\Delta \nu##.

 

[gboff21]

E=hc/λ
and either way (nu or lambda) the answer is stupidly large

 

[mfb]

E=hc/λ, but not ΔE=hc/Δλ!
$$E+\Delta E=\frac{hc}{\lambda+\Delta \lambda} \approx \frac{hc}{\lambda}(1-\frac{\Delta \lambda}{\lambda})$$
Simplified:
$$\Delta E=-\Delta \lambda \frac{hc}{\lambda^2}$$

It is easier to use relative values everywhere:
$$\frac{\Delta E}{E}=\frac{\Delta \nu}{\nu}=-\frac{\Delta \lambda}{\lambda}$$

 

[gboff21]

Now it's so obvious!
Thanks a lot mfb!