Spectral Line Width and Uncertainty principle

Question

so I've been at this for about 3 - 4 hours now. It is an homework assignment (well part of a question which i've already completed). We did not learn this in class. All work is shown below.

An atom in an excited state of $4.9 eV$ emits a photon and ends up in the ground state. The lifetime of the excited state is $1.2 \times 10^{-13} s$.

(b) What is the spectral line width (in wavelength) of the photon?

So lets look at what I have done so far. I have done the following: $$\Delta E \Delta t = \frac{\hbar}{2} $$ but $$E = h f$$ so $$\Delta f = \frac{1}{4\pi \Delta t}$$

but if I take $\Delta f$ and convert it into wavelength using $\lambda f = c $ then it gives me the wrong answer. I've tried MANY variations of the above formulas.

The correct answer is $0.142 nm $

Can anyone give me a hint?

Answer 1

Hint: Your problem is in the "take $\Delta f$ an convert it to wavelength using $\lambda f = c$" part. The equation $\lambda f = c$ does not imply $\Delta \lambda \Delta f = c$.

Answer: Rather it implies,

$\lambda f = c$

$\lambda = \frac{c}{f} $

Now differentiate: $d\lambda = -c\frac{df}{f^2}$

$df \ll f$ so treating the $\Delta f$ as a differential works fine. If this was not the case, you'd want to integrate from $f_{min}$ to $f_{max}$ (not given in the problem, just how you'd have to do it otherwise). This formula gives you the right answer. The sign opposite signs just indicate the higher frequency corresponds to the lower wavelength and is ignored in the final answer.

Answer 2

Okay so a buddy helped me out.

You had to use the following formula:

$$ \Delta \lambda = hc \( \frac{1}{E_1} - \frac{1}{E_2} \) $$