# Homework Help: Linewidth of emission and the uncertainty principle

1. Oct 11, 2011

### Kweh-chan

1. The problem statement, all variables and given/known data
The linewidth of a certain fluorescing dye molecule is 60nm and the peak emission wavelength is 600nm. The lifetime of the dye molecule in the excited state is 1ns. Determine whether the linewidth of the emission is a result of the uncertainty principle.
(it may be helpful to convert wavelengths into appropriate frequency units)

2. Relevant equations
ΔEΔt ≥ h/2pi
λ = c/v

3. The attempt at a solution
I solved for energy in joules
ΔE ≥ h/2pi* Δt
ΔE ≥ (6.626x10^-34 Js)/(2pi * (1.0x10^-9s)) = 1.05x10^-25 J

and in eV
1J = 6.25x10^18 ev
(1.05x10^-25 J) * (6.25x10^18) = 6.56x10^-7 eV

I've also converted 600x10^-9m into a frequency of v = 5x10^14
I haven't converted the 60nm as it says it's a linewidth, which I am assuming is a distance, rather than a wavelength.

I have some notes about a similar problem from last year, and after determining energy, it calculates "fractional uncertainty in energy," and since there are no units in these notes (which were printed and provided for us), I am not entirely certain how to calculate this or if it is even relevant to the problem at hand.

The notes then go onto a calculation which does as follows
λ=hc/E --> dλ = (-hc/E^2) * dE = -(hc/E)*(dE/E) = λ*(frac. unc. in energy)
According to this, this will get me to fractional uncertainty in wavelength. I understand this equation, but just not how to get to this point, or, again, if it is even relevant.

I have tried putting these two piece of information together to help me get to my answer, but I cannot quite seem to make a connection. I am guessing I will use the energy I calculated to deduce another wavelength value and use that to compare to the one given to me, in order to determine if the one given comes from the uncertainty...but I still am confused about where the linewidth then comes in.

2. Oct 11, 2011

### gash789

Personally I would take this as a hint and begin here,

then consider the equation they have given you ΔEΔt ≥ h/2pi

You have now, the Δt, but not an energy.

But you do have a frequency...