# Derive a formula from the uncertainty principle

1. Jan 20, 2014

### bobby.pdx

1. The problem statement, all variables and given/known data
Derive from the uncertainty principle a formula for the relative spread of the spectral line that corresponds to the longest wavelength of the Lyman series.

2. Relevant equations
uncertainty principle:
σxσp≥$\hbar$/2

planck constant
$\hbar$=h/2pi
h=λp

Lyman series:
1/λ=RH(1-1/n2)

λ=hc/Ei-Ef

3. The attempt at a solution
I'm not quite sure how to go about the problem. I have gathered some formulas I believe will help me out. If I substitute some of these formulas in to the uncertainty principle I get

σxσp≥(hc/Ei-Ef)p/(4pi)

I'm not sure where to go from here. Any help would be greatly appreciated.

2. Jan 21, 2014

### hilbert2

I don't think it's possible to solve the problem using only those equations. You are probably expected to approximate the "lifetime broadening" of the spectral line. You'd have to know the lifetime/transition rate of the excited state corresponding to the spectral line, and use the time-energy uncertainty relation $\Delta E \Delta t \geq \hbar / 2$.

3. Jan 21, 2014

### bobby.pdx

Let's say the lifetime of the excited state is 10^-7 seconds. How would I go about deriving the formula from there?

4. Jan 21, 2014

### hilbert2

Well, if you know the lifetime $\Delta t$, then the spectral linewidth is just $\Delta E \approx \frac{\hbar}{2\Delta t}$.

5. Jan 21, 2014

### bobby.pdx

This seems right. The only thing is the problem then asks to use this formula to calculate this kind of spread of spectral lines for both hydrogen and tritium for this spectral line with and without the reduced mass correction to the Bohr model of both hydrogen and tritium. If this formula is correct then the answer would be the same for both hydrogen and tritium right?