# Natural broadening of lithium sub-levels

1. Nov 11, 2012

### gboff21

1. The problem statement, all variables and given/known data
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 $\frac{\Delta \lambda}{\lambda} ~= \frac{\Delta \nu}{\nu}$ at $\frac{\Delta \nu}{\nu} << 1$

2. Relevant equations

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

3. The attempt at a solution
uncertainty principle: $\Delta E \tau = \hbar$ (tau is lifetime)
so $\Delta \lambda = 2 \pi c \tau$ (which comes out as ~50m)
This is clearly wrong since the doppler broadening comes out as in the order of 10^-12.

Any clues?

2. Nov 11, 2012

### Staff: Mentor

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 $\Delta \nu$.

3. Nov 11, 2012

### gboff21

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

4. Nov 11, 2012

### Staff: Mentor

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:
$$\frac{\Delta E}{E}=\frac{\Delta \nu}{\nu}=-\frac{\Delta \lambda}{\lambda}$$

5. Nov 11, 2012

### gboff21

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